In a hydraulic lift, the surface area of the input piston is 6 cm2 and that of the output piston is 1500 cm2. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is _______ kJ.
Explanation:
$\begin{aligned} & \frac{\mathrm{F}_1}{\mathrm{~A}_1}=\frac{\mathrm{F}_2}{A_2}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} \times 1500 \\ & \mathrm{~F}=50 \times 500=25 \times 10^3 \mathrm{~N} \\ & \omega=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{~S}}=25 \times 10^3 \times \frac{20}{100} \\ & =5 \times 10^3=5 \mathrm{~kJ} \end{aligned}$
A force $\mathrm{f}=\mathrm{x}^2 \mathrm{y} \hat{\mathrm{i}}+\mathrm{y}^2 \hat{\mathrm{j}}$ acts on a particle in a plane $\mathrm{x}+\mathrm{y}=10$. The work done by this force during a displacement from $(0,0)$ to $(4 \mathrm{~m}, 2 \mathrm{~m})$ is _________ Joule (round off to the nearest integer)
Explanation:
To calculate the work done by the force $\mathrm{f} = \mathrm{x}^2 \mathrm{y} \hat{\mathrm{i}} + \mathrm{y}^2 \hat{\mathrm{j}}$ during the displacement from $(0,0)$ to $(4 \text{ m}, 2 \text{ m})$, we proceed as follows:
Given the constraint that the particle moves in the plane $\mathrm{x} + \mathrm{y} = 10$, we can express $\mathrm{y}$ in terms of $\mathrm{x}$:
$ y = 10 - x $
The work done, $W$, is the integral of the force along the path of the displacement:
$ W = \int_{0}^{4} x^2 (10-x) \, dx + \int_{0}^{2} y^2 \, dy $
Calculate the integral of the $\hat{\mathrm{i}}$ component:
$ \begin{aligned} & \int_{0}^{4} x^2 (10 - x) \, dx = \int_{0}^{4} (10x^2 - x^3) \, dx \\ & = \left[ \frac{10x^3}{3} - \frac{x^4}{4} \right]_{0}^{4} \\ & = \left( \frac{10(4)^3}{3} - \frac{(4)^4}{4} \right) - \left( \frac{10(0)^3}{3} - \frac{(0)^4}{4} \right) \\ & = \frac{640}{3} - \frac{256}{4} \\ \end{aligned} $
Calculate the integral of the $\hat{\mathrm{j}}$ component:
$ \begin{aligned} & \int_{0}^{2} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{2} \\ & = \frac{(2)^3}{3} - \frac{(0)^3}{3}\\ & = \frac{8}{3} \\ \end{aligned} $
Combine the results:
$ W = \left( \frac{640}{3} - 64 \right) + \frac{8}{3} $
Simplify:
$ \begin{aligned} W & = \frac{640}{3} - 64 + \frac{8}{3} \\ & = \frac{640 + 8}{3} - 64 \\ & = \frac{648}{3} - 64 \\ & = 216 - 64 \\ & = 152 \text{ J} \end{aligned} $
Thus, the work done by the force is 152 Joules.
A force $(3 x^2+2 x-5) \mathrm{N}$ displaces a body from $x=2 \mathrm{~m}$ to $x=4 \mathrm{~m}$. Work done by this force is ________ J.
Explanation:
To find the work done by the force when the body is displaced from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$, we use the formula for work done by a variable force in one dimension, which is the integral of the force with respect to displacement:
$ W = \int_{x_1}^{x_2} F \, dx $
Given the force $F(x) = (3x^2 + 2x - 5) \, \mathrm{N}$ and the limits of integration from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$, we can substitute these values into the equation:
$ W = \int_{2}^{4} (3x^2 + 2x - 5) \, dx $
Calculating the integral, we get:
$ W = \left[\frac{3x^3}{3} + \frac{2x^2}{2} - 5x\right]_2^4 $
This simplifies to:
$ W = \left[x^3 + x^2 - 5x\right]_2^4 $
Substituting the upper limit ($x = 4$) and then the lower limit ($x = 2$) into the antiderivative, and subtracting the latter from the former, we get:
$ W = \left[(4)^3 + (4)^2 - 5(4)\right] - \left[(2)^3 + (2)^2 - 5(2)\right] $
$ W = (64 + 16 - 20) - (8 + 4 - 10) $
$ W = 60 - 2 $
$ W = 58 \, \mathrm{J} $
Therefore, the work done by the force as the body displaces from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$ is $58 \, \mathrm{J}$.
Explanation:
$W = \int_{x_1}^{x_2} F(x) dx$
In this case, $F(x) = 5x$, $x_1 = 2m$, and $x_2 = 4m$. Now, we can substitute these values into the formula and evaluate the integral:
$W = \int_{2}^{4} 5x dx$
To evaluate the integral, we find the antiderivative of $5x$:
$\int 5x dx = \frac{5}{2}x^2 + C$
Now, we can find the work done by evaluating the antiderivative at the limits of integration:
$W = \left[\frac{5}{2}x^2\right]_{2}^{4} = \frac{5}{2}(4^2) - \frac{5}{2}(2^2)$
$W = \frac{5}{2}(16) - \frac{5}{2}(4) = 40 - 10 = 30 \mathrm{J}$
The work done by the force in moving the block from $x = 2m$ to $x = 4m$ is 30 J.
A car accelerates from rest to $u \mathrm{~m} / \mathrm{s}$. The energy spent in this process is E J. The energy required to accelerate the car from $u \mathrm{~m} / \mathrm{s}$ to $2 \mathrm{u} \mathrm{m} / \mathrm{s}$ is $\mathrm{nE~J}$. The value of $\mathrm{n}$ is ____________.
Explanation:
$K = \frac{1}{2}mv^2$
The work done in accelerating an object from rest to velocity $v$ is equal to its change in kinetic energy. Therefore, the energy spent in accelerating the car from rest to $u \mathrm{~m}/\mathrm{s}$ is:
$E = \frac{1}{2}mu^2$
The energy required to accelerate the car from $u \mathrm{~m}/\mathrm{s}$ to $2u \mathrm{~m}/\mathrm{s}$ is:
$\begin{aligned} nE &= \frac{1}{2}m(2u)^2 - \frac{1}{2}mu^2 \\\\ &= 2mu^2 - \frac{1}{2}mu^2 \\\\ &= \frac{3}{2}mu^2 \\\\ &= 3E \end{aligned} $
$ \therefore $ n = 3
To maintain a speed of 80 km/h by a bus of mass 500 kg on a plane rough road for 4 km distance, the work done by the engine of the bus will be ____________ KJ. [The coefficient of friction between tyre of bus and road is 0.04.]
Explanation:
$F_{friction} = \mu F_N$
Where $\mu$ is the coefficient of friction and $F_N$ is the normal force acting on the bus. Since the bus is on a flat road, the normal force is equal to the gravitational force:
$F_N = mg$
Where $m$ is the mass of the bus and $g$ is the acceleration due to gravity (approximately $9.8 \mathrm{~m/s^2}$).
Substituting the values, we get:
$F_{friction} = 0.04 \times 500 \times 9.8$
$F_{friction} = 196 \mathrm{~N}$
To maintain a constant speed, the engine must exert a force equal in magnitude to the frictional force. The work done by the engine to overcome the frictional force is given by:
$W = F_{friction} \times d$
Where $d$ is the distance traveled. First, convert the distance from km to m:
$d = 4 \mathrm{~km} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} = 4000 \mathrm{~m}$
Now, calculate the work done:
$W = 196 \mathrm{~N} \times 4000 \mathrm{~m}$ $W = 784000 \mathrm{~J}$
Convert the work done from joules to kilojoules:
$W = \frac{784000 \mathrm{~J}}{1000 \mathrm{~J/ kJ}} = 784 \mathrm{~kJ}$
The work done by the engine of the bus to maintain a speed of 80 km/h for a 4 km distance is $784.8 \mathrm{~kJ}$.
A block of mass $5 \mathrm{~kg}$ starting from rest pulled up on a smooth incline plane making an angle of $30^{\circ}$ with horizontal with an affective acceleration of $1 \mathrm{~ms}^{-2}$. The power delivered by the pulling force at $t=10 \mathrm{~s}$ from the start is ___________ W.
[use $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]
(calculate the nearest integer value)
Explanation:
To find the power delivered by the pulling force at t = 10 s, we first need to find the work done by the force. The work done is given by the product of force and displacement, and the power is the rate of work done.
Calculate the velocity (v) at t = 10 s:
Since the block starts from rest and is pulled up with an effective acceleration of 1 m/s², we can use the equation of motion to find the velocity (v) at t = 10 s:
$v = u + at$
Here, u = 0 (initial velocity) and a = 1 m/s² (acceleration). Plugging in the values:
$v_{10} = 0 + 1(10) = 10 \mathrm{~m/s}$
- Calculate the net force acting on the block ($F_{net}$):
The net force acting on the block along the incline plane is the difference between the pulling force (F) and the gravitational force component acting parallel to the incline (mgsinθ):
$F_\text{net} = F - mgsinθ$
Since F_net = ma, we can write:
$F = ma + mgsinθ$
Plugging in the values (m = 5 kg, a = 1 m/s², g = 10 m/s², and θ = 30°):
$F = 5(1) + 5(10)(\sin 30°) = 5 + 25 = 30 \mathrm{~N}$
Calculate the power (P) at t = 10 s:
The power (P) can be calculated as the product of force (F) and velocity (v):
$P_{10} = Fv = 30(10) = 300 \mathrm{~W}$
So, the power delivered by the pulling force at t = 10 s from the start is 300 W.
A force $\vec{F}=(2+3 x) \hat{i}$ acts on a particle in the $x$ direction where F is in newton and $x$ is in meter. The work done by this force during a displacement from $x=0$ to $x=4 \mathrm{~m}$, is __________ J.
Explanation:
To find the work done by a force during a displacement, we can use the formula:
$W = \int_{x_1}^{x_2} \vec{F} \cdot d\vec{x}$
Here, the force is given by $\vec{F} = (2+3x) \hat{i}$, and we need to find the work done during a displacement from $x = 0$ to $x = 4 \mathrm{~m}$. Since the force is only in the $x$ direction, we can write the integral as:
$W = \int_{0}^{4} (2+3x) dx$
Now we can integrate the function with respect to $x$:
$W = \int_{0}^{4} (2+3x) dx = \int_{0}^{4} 2 dx + \int_{0}^{4} 3x dx$
$W = \left[ 2x \right]_0^4 + \left[ \frac{3}{2}x^2 \right]_0^4$
Now we can plug in the limits of integration:
$W = (2 \cdot 4 - 2 \cdot 0) + \left(\frac{3}{2} \cdot 4^2 - \frac{3}{2} \cdot 0^2 \right)$
$W = 8 + 24$
$W = 32 \mathrm{~J}$
So the work done by the force during the displacement from $x = 0$ to $x = 4 \mathrm{~m}$ is 32 Joules.
If the maximum load carried by an elevator is $1400 \mathrm{~kg}$ ( $600 \mathrm{~kg}$ - Passengers + 800 $\mathrm{kg}$ - elevator), which is moving up with a uniform speed of $3 \mathrm{~m} \mathrm{~s}^{-1}$ and the frictional force acting on it is $2000 \mathrm{~N}$, then the maximum power used by the motor is __________ $\mathrm{kW}\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
Explanation:
First, let's find the total weight of the elevator and passengers:
Total weight = (mass of passengers + mass of elevator) × g
Total weight = (600 kg + 800 kg) × 10 m/s²
Total weight = 1400 kg × 10 m/s² = 14,000 N
Now, we need to calculate the total force acting on the elevator as it moves upwards. Since the elevator is moving at a constant speed, the net force acting on it is zero. Therefore, the tension in the cable must balance the total weight and frictional force:
Tension = Total weight + Frictional force Tension = 14,000 N + 2,000 N = 16,000 N
The power used by the motor can be calculated using the formula:
Power = Force × Velocity
Here, the force is the tension in the cable, and the velocity is the speed of the elevator:
Power = 16,000 N × 3 m/s = 48,000 W
To convert the power to kilowatts, divide by 1,000:
Power = 48,000 W / 1,000 = 48 kW
So, the maximum power used by the motor is 48 kW.
A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in vertical plane. A block of mass 1 kg just fit inside the tube. The speed of block is 22 m/s, when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The work done by the tube on the block is __________ J. (Given g = 10 m/s$^2$).
Explanation:
$ \begin{aligned} &W_{\text {gravity }}+W_{\text {friction }} =\Delta(K E)=K E_f-K E_i \\\\ &W_{\text {gravity }} =m g h=1 \times 10 \times 0.3=3 \mathrm{~J} \\\\ &W_{\text {friction }} =0-\frac{1}{2} \times(22)^2-3 \\\\ & =-(242+3)=-245 \mathrm{~J} \end{aligned} $
The negative sign indicates that the work done by frictional force (the tube) is in the direction opposite to the displacement of the block. In conclusion, the work done by the tube on the block is 245 J.
A body of mass $5 \mathrm{~kg}$ is moving with a momentum of $10 \mathrm{~kg} \mathrm{~ms}^{-1}$. Now a force of $2 \mathrm{~N}$ acts on the body in the direction of its motion for $5 \mathrm{~s}$. The increase in the Kinetic energy of the body is ___________ $\mathrm{J}$.
Explanation:
The increase in kinetic energy can be found using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
The work done by a force is given by the equation:
$ W = F \cdot d $
where ( F ) is the force and ( d ) is the distance over which the force is applied.
However, we don't have the distance in this problem. But we do know that the force is applied for a time of 5 seconds, and that the initial momentum of the body is 10 kg m/s. We can use these facts to find the work done.
First, we can use the equation for force, ( F = ma ), to find the acceleration of the body:
$a = \frac{F}{m} = \frac{2 \, \text{N}}{5 \, \text{kg}} = 0.4 \, \text{m/s}^2 $
Then, we can use the equation for distance in uniformly accelerated motion, ( $d = v_i t + \frac{1}{2} a t^2 $), where ( $v_i$ ) is the initial velocity of the body. We can find ( $v_i $) from the initial momentum and the mass of the body:
$ v_i = \frac{p}{m} = \frac{10 \, \text{kg m/s}}{5 \, \text{kg}} = 2 \, \text{m/s} $
Substituting ( $v_i$ ), ( a ), and ( t ) into the equation for ( d ) gives:
$ d = 2 \, \text{m/s} \cdot 5 \, \text{s} + \frac{1}{2} \cdot 0.4 \, \text{m/s}^2 \cdot (5 \, \text{s})^2 = 10 \, \text{m} + 5 \, \text{m} = 15 \, \text{m} $
Finally, we can substitute ( F ) and ( d ) into the equation for work to find the increase in kinetic energy:
$ \Delta KE = W = F \cdot d = 2 \, \text{N} \cdot 15 \, \text{m} = 30 \, \text{J} $
So, the increase in the kinetic energy of the body is 30 J.
A body is dropped on ground from a height '$h_{1}$' and after hitting the ground, it rebounds to a height '$h_{2}$'. If the ratio of velocities of the body just before and after hitting ground is 4 , then percentage loss in kinetic energy of the body is $\frac{x}{4}$. The value of $x$ is ____________.
Explanation:
The velocity of the body just before hitting the ground, due to gravitational acceleration, is given by $v_{1} = \sqrt{2gh_{1}}$, and the velocity just after hitting the ground, when it rebounds to a height $h_{2}$, is given by $v_{2} = \sqrt{2gh_{2}}$.
According to the problem, the ratio $\frac{v_{1}}{v{2}} = 4$. Therefore, we can write $\frac{\sqrt{2gh_{1}}}{\sqrt{2gh_{2}}} = 4$ or equivalently $\frac{h_{1}}{h_{2}} = 4^2 = 16$.
The loss in kinetic energy due to the collision with the ground is given by the difference between the initial kinetic energy $K_{1} = \frac{1}{2} m v_{1}^2$ and the final kinetic energy $K_{2} = \frac{1}{2} m v_{2}^2$, where m is the mass of the body.
Substituting $v_{1} = \sqrt{2gh_{1}}$ and $v_{2} = \sqrt{2gh_{2}}$ into these expressions, we get $K_{1} = mgh_{1}$ and $K_{2} = mgh_{2}$.
The loss in kinetic energy is then $\Delta K = K_{1} - K_{2} = mgh_{1} - mgh_{2}$.
The percentage loss in kinetic energy is given by
$\frac{\Delta K}{K_{1}} \times 100 = \frac{mgh_{1} - mgh_{2}}{mgh_{1}} \times 100 = \frac{h_{1} - h_{2}}{h_{1}} \times 100$.
Since $h_{1}/h_{2} = 16$, we can write $h_{2} = h_{1}/16$, so the percentage loss in kinetic energy is
$\frac{h_{1} - h_{1}/16}{h_{1}} \times 100 = 100(1 - \frac{1}{16}) = 100 \times \frac{15}{16} = \frac{375}{4}$.
So, the value of $x$ is 375.
A particle of mass $10 \mathrm{~g}$ moves in a straight line with retardation $2 x$, where $x$ is the displacement in SI units. Its loss of kinetic energy for above displacement is $\left(\frac{10}{x}\right)^{-n}$ J. The value of $\mathrm{n}$ will be __________
Explanation:
The work done against the retarding force is indeed equal to the loss in kinetic energy.
The force acting on the particle due to retardation is given by $F = ma = -2mx$.
When we integrate this force over the displacement from $0$ to $x$, we get:
$\Delta KE = W = \int F \cdot dx = \int (-2mx) \, dx = -mx^2$
The negative sign indicates that this is a loss of kinetic energy.
The problem states that the loss in kinetic energy is also given by $\left(\frac{10}{x}\right)^{-n}$ J. Therefore, we have:
$-mx^2 = \left(\frac{10}{x}\right)^{-n}$
Because this is a loss of kinetic energy, we should consider the absolute value. Hence,
$mx^2 = \left(\frac{10}{x}\right)^{-n}$
Substituting the given mass $m = 10 \, \text{g} = 0.01 \, \text{kg}$, we get:
$0.01x^2 = \left(\frac{10}{x}\right)^{-n}$
This simplifies to:
$x^2 = \left(\frac{10}{x}\right)^{-n}$
Comparing the two sides, we can see that $n = 2$.
A block is fastened to a horizontal spring. The block is pulled to a distance $x=10 \mathrm{~cm}$ from its equilibrium position (at $x=0$) on a frictionless surface from rest. The energy of the block at $x=5$ $\mathrm{cm}$ is $0.25 \mathrm{~J}$. The spring constant of the spring is ___________ $\mathrm{Nm}^{-1}$
Explanation:
Spring energy at x = 10 cm,
$\mathrm{U}_{\mathrm{i}} =\frac{1}{2} \mathrm{kx}_0^2 $
Energy of the block at x = 10,
$\mathrm{~K}_{\mathrm{i}} =0$
Spring energy at x = 5 cm,
$\mathrm{U}_{\mathrm{f}}=\frac{1}{2} \mathrm{k}\left(\frac{\mathrm{x}_0}{2}\right)^2 $
Energy of the block at x = 5, (which is only kinetic energy, no potential energy of block presents as block is not moving in the vertical direction)
$ \mathrm{~K}_{\mathrm{f}}=0.25 \mathrm{~J} $
Applying energy conservation law,
Initial energy of Spring + Initial energy of Block = Final energy of Spring + Final energy of Block
$ \frac{1}{2} \mathrm{kx}_0^2+0=\frac{1}{2} \mathrm{k} \frac{\mathrm{x}_0^2}{4}+0.25 $
$ \frac{1}{2} \mathrm{kx}_0^2 \frac{3}{4}=\frac{1}{4} $
$ \frac{1}{2} \mathrm{k} \frac{3}{100}=1 \Rightarrow \mathrm{k}=\frac{200}{3} \mathrm{~N} / \mathrm{m} $
$ =67 \mathrm{~N} / \mathrm{m} $
A force $\mathrm{F}=\left(5+3 y^{2}\right)$ acts on a particle in the $y$-direction, where $\mathrm{F}$ is in newton and $y$ is in meter. The work done by the force during a displacement from $y=2 \mathrm{~m}$ to $y=5 \mathrm{~m}$ is ___________ J.
Explanation:
A small particle moves to position $5 \hat{i}-2 \hat{j}+\hat{k}$ from its initial position $2 \hat{i}+3 \hat{j}-4 \hat{k}$ under the action of force $5 \hat{i}+2 \hat{j}+7 \hat{k} \mathrm{~N}$. The value of work done will be __________ J.
Explanation:
To find the work done, we use the dot product of the force and displacement vectors :
$ \begin{aligned} & W=\vec{F} \cdot\left(\vec{r}_f-\vec{r}_{\mathrm{i}}\right) \\\\ & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot((5 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k})) \\\\ & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}+5 \hat{k}) \\\\ & =15-10+35 \\\\ & =40 \mathrm{~J} \end{aligned} $
A lift of mass $\mathrm{M}=500 \mathrm{~kg}$ is descending with speed of $2 \mathrm{~ms}^{-1}$. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2 \mathrm{~ms}^{-2}$. The kinetic energy of the lift at the end of fall through to a distance of $6 \mathrm{~m}$ will be _____________ $\mathrm{kJ}$.
Explanation:
$ \begin{aligned} & a=2 \mathrm{~m} / \mathrm{s}^{2} \\\\ & s=6 \mathrm{~m} \\\\ & v=? \\\\ & v^{2}=u^{2}+2 a s \\\\ & v^{2}=4+2 \times 2 \times 6 \\\\ & =28 \end{aligned} $
So, $\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 500 \times 28 \mathrm{~J}$
$=7000 \mathrm{~J}$
$=7 \mathrm{~kJ}$
Explanation:
$P = Fv$
$m{{vdv} \over {dt}} = P$
$m\int_0^v {vdv = \int_0^t {Pdt} } $
${{m{v^2}} \over 2} = Pt$
$v = \sqrt {{{2P} \over m}} {t^{1/2}}$
$\int_0^s {dx = \sqrt {{{2P} \over m}} \int_0^t {{t^{1/2}}dt} } $
$s = {2 \over 3}\sqrt {{{2P} \over m}} {t^{3/2}}$
or $s = {2 \over 3}\sqrt {{{2P} \over 2}} \times {4^{3/2}}$
$ = {{16} \over 3}\sqrt P ~m$
So, $\alpha = 4$
A 0.4 kg mass takes 8s to reach ground when dropped from a certain height 'P' above surface of earth. The loss of potential energy in the last second of fall is __________ J.
(Take g = 10 m/s$^2$)
Explanation:
$\mathrm{S}_{8}=0+\frac{1}{2} \times 10 \times(2 \times 8-1)$
$\mathbf{S}_{8}=5 \times 15$
$\Delta \mathrm{U}=0.4 \times 10 \times 5 \times 15$
$\Delta \mathrm{U}=20 \times 15=300$
An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F = 2N. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta=\mathrm{k}x$, where k is a constant and $x$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be $E = {n \over k}\sin \theta $. The value of n is ___________.
Explanation:
$ \begin{aligned} & \therefore \int F \cdot d x=\frac{1}{2} m v^{2}=E \\\\ & \therefore E=\int_{0}^{x} 2 \cos (k x) d x \\\\ & E=\frac{2}{k}[\sin k x]_{0}^{x} \\\\ & =\frac{2}{k} \sin k x \\\\ & =\frac{2 \sin \theta}{k} \end{aligned} $
A body of mass 1kg begins to move under the action of a time dependent force $\overrightarrow F = \left( {t\widehat i + 3{t^2}\,\widehat j} \right)$ N, where $\widehat i$ and $\widehat j$ are the unit vectors along $x$ and $y$ axis. The power developed by above force, at the time t = 2s, will be ____________ W.
Explanation:
A spherical body of mass 2 kg starting from rest acquires a kinetic energy of 10000 J at the end of $\mathrm{5^{th}}$ second. The force acted on the body is ________ N.
Explanation:
So displacement $S=\frac{1}{2} a t^{2}=\frac{F t^{2}}{2 m}$
So work done $W=F . S=\frac{F^{2} t^{2}}{2 m}$
From work energy Theorem
$\Delta K E=W$
$W=\frac{F^{2} t^{2}}{2 m}=10000$
$F=\sqrt{\frac{10000 \times 2 \times 2}{5^{2}}}$
$F=40 \mathrm{~N}$
A block of mass '$\mathrm{m}$' (as shown in figure) moving with kinetic energy E compresses a spring through a distance $25 \mathrm{~cm}$ when, its speed is halved. The value of spring constant of used spring will be $\mathrm{nE} \,\,\mathrm{Nm}^{-1}$ for $\mathrm{n}=$ _________.
Explanation:
$\Delta KE = {W_{all}}$
So ${E \over 4} - E = - {1 \over 2}K \times {(0.25)^2}$
${{3E} \over 4} = {1 \over 2}K \times {1 \over {16}}$
$ = K = 24E$
A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is ___________ m.
Explanation:
$(x)g\lambda = \mu (6 - x)\,g\lambda $ where x is length of hanging part
$ \Rightarrow x = 3 - 0.5x$
$ \Rightarrow x = 2$ m
A 0.5 kg block moving at a speed of 12 ms$-$1 compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be _______________ Nm$-$1.
Explanation:
${1 \over 2}m\,{V^2} = {1 \over 2}k{x^2} + {1 \over 2}m{\left( {{v \over 2}} \right)^2}$
$ \Rightarrow {3 \over 8}m{v^2} = {1 \over 2}k{x^2}$
$ \Rightarrow k = {3 \over 4} \times {1 \over 2} \times {{144} \over 9} \times 100$
$ = 600$
$ \Rightarrow 600$
A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance ${h \over 2}$. The spring constant is _____________ Nm$-$1.
(Use g = 10 ms$-$2)
Explanation:
$mg\left( {h + {h \over 2}} \right) = {1 \over 2}k{\left( {{h \over 2}} \right)^2}$
$ \Rightarrow 0.1 \times 10\times(0.15) = {1 \over 2}k{(0.05)^2}$
$ \Rightarrow k = 120$ N/m
Explanation:
The total mass of the system, M = 40000 kg
The speed of the wagon, v = 72 km/h
When brakes are applied, the final velocity, vf = 0
The compressed spring of the shock absorber, x = 1 m
Applying the work-energy theorem,
Work done by the system = Change in kinetic energy
$W = \Delta KE$
${W_{friction}} + {W_{spring}} = {1 \over 2}mv_f^2 + {1 \over 2}mv_i^2$
$ - {{90} \over {100}}\left( {{1 \over 2}m{v^2}} \right) + {W_{spring}} = 0 - {1 \over 2}mv_i^2$ ($\because$ 90% energy lost due to friction)
${W_{spring}} = - {{10} \over {100}} \times {1 \over 2}m{v^2}$
$ - {1 \over 2}k{x^2} = {1 \over {20}}m{v^2}$
$k = {{m{v^2}} \over {10 \times {x^2}}}$
Substituting the values in the above equation, we get
$k = {{40000 \times {{\left( {72 \times {5 \over {18}}} \right)}^2}} \over {10{{(1)}^2}}}$
= 16 $\times$ 105 N/m
Comparing the spring constant, k = x $\times$ 105
The value of the x = 16.
Explanation:
Pi = Pf
m $\times$ 40 = ${m \over 2}$ $\times$ v + ${m \over 2}$ $\times$ 60
40 = ${v \over 2}$ + 30
$\Rightarrow$ v = 20
(K. E.)I = ${1 \over 2}$m $\times$ (40)2 = 800 m
(K. E.)f = ${1 \over 2}$${m \over 2}$ . (20)2 + ${1 \over 2}$ . ${m \over 2}$ (60)2 = 1000 m
| $\Delta$ K. E. | = | 1000m $-$ 800 m | = 200 m
${{\Delta K.E.} \over {{{(K.E.)}_i}}} = {{200m} \over {800m}} = {1 \over 4} = {x \over 4}$
x = 1
Explanation:
FAd cos45$^\circ$ = FBd cos60$^\circ$
${F_A} \times {1 \over {\sqrt 2 }} = {F_B} \times {1 \over 2}$
${{{F_A}} \over {{F_B}}} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}$
x = 2
Explanation:
From energy conservation
Ki + Ui = kf + Uf
$0 + \left( { - 1 \times 10 \times {1 \over 2}} \right) = {k_f} + \left( { - 3 \times 10 \times {3 \over 2}} \right)$
$-$5 = kf $-$ 45
kf = 40 J
(Assume there is no friction between the block and the hemisphere)
Explanation:
$mg\cos \theta = {{m{v^2}} \over R}$ .... (1)
$\cos \theta = {h \over R}$
Energy conservation
$mg\{ R - h\} = {1 \over 2}m{v^2}$ ..... (2)
from (1) & (2)
$\Rightarrow$ $mg\left\{ {{h \over R}} \right\} = {{2mg\{ R - h\} } \over R}$
$h = {{2R} \over 3}$ = 2m
Explanation:
$W = \int {Fdy = \int\limits_0^{10} {(5y + 20)dy} } $
$ = \left( {{{5{y^2}} \over 2} + 20y} \right)_0^{10}$
$ = {5 \over 2} \times 100 + 20 \times 10$
$ = 250 + 200 = 450$ J
Explanation:
m = 100 g = 0.1 kg
x = 0.05 m and H = 2 m
By energy conservation,
${1 \over 2}k{x^2} = {1 \over 2}m{v^2} \Rightarrow v = x\sqrt {{k \over m}} $
$ = 0.05 \times \sqrt {{{100} \over {0.1}}} = 0.5\sqrt {10} $ ms$-$1 .... (i)
Time of flight of ball, $t = \sqrt {{{2H} \over g}} $
$ \Rightarrow t = \sqrt {{{2 \times 2} \over {10}}} = {2 \over {\sqrt {10} }}$s .... (ii)
$\therefore$ Range of ball, d = vt
$ = 0.5\sqrt {10} \times \left( {{2 \over {\sqrt {10} }}} \right)$ [From Eqs. (i) and (ii)]
$ = 0.5 \times 2 = 1 m$
Explanation:
If spring compressed by x,
then work done by spring = 0 $-$ ${1 \over 2}$ $\times$ 4 $\times$ 102
Applying work energy theorem,
$-$${1 \over 2}$ kx2 = $-$${1 \over 2}$ $\times$ 4 $\times$ 102
$ \Rightarrow $ 100x2 = 4 $\times$ 102
$ \Rightarrow $ x = 2
$ \therefore $ Final length of the spring = 8 $-$ 2 = 6 m
The value of 'x' to the nearest integer is __________.
Explanation:
Ki + Ui = Kf + Uf
0 + 10 $\times$ 10 $\times$ 10 = ${1 \over 2}$ $\times$ 10 $\times$ v$_B^2$ + 10 $\times$ 10 $\times$ 5
1000 = 5v$_B^2$ + 500
v$_B^2$ = ${{500} \over 5}$ = 100
VB = 10 m/s
x = 10
$U = {\alpha \over {{r^{10}}}} - {\beta \over {{r^5}}} - 3$
where, $\alpha$ and $\beta$ are positive constants. The equilibrium distance between two atoms will be ${\left( {{{2\alpha } \over \beta }} \right)^{{a \over b}}}$, where a = ___________.
Explanation:
$F = - \left[ { - {{10\alpha } \over {{r^{11}}}} + {{5\beta } \over {{r^6}}}} \right]$
for equilibrium, F = 0
${{10\alpha } \over {{r^{11}}}} = {{5\beta } \over {{r^6}}}$
${{2\alpha } \over \beta } = {r^5}$
$r = {\left( {{{2\alpha } \over \beta }} \right)^{1/5}}$
$ \therefore $ $a = 1$
Explanation:
Let s be the required distance.
From Work - Energy theorem,
Work = Change in kinetic energy
$\Rightarrow$ Power $\times$ Time = $\Delta$K
i.e., Pt = $\Delta$K $\Rightarrow$ Pt = ${1 \over 2}$mv2 ..... (i)
Given, P = 1 Js$-$1, t = 9 s, m = 2 kg
Substituting all the given values in eq. (i), we get
1 $\times$ 9 = ${1 \over 2}$(2) v2
v2 = 9 $\Rightarrow$ v = 3 m/s (at t = 9 s)
As, Fv = P $\Rightarrow$ (ma)v = P [$\because$ F = ma]
$ \Rightarrow m\left[ {{{dv} \over {dt}}} \right]v = P \Rightarrow m\left[ {{{ds} \over {dt}}{{dv} \over {ds}}} \right]v = P$
$ \Rightarrow m\left[ {v{{dv} \over {ds}}} \right]v = P$
$ \Rightarrow 2{v^2}dv = ds$ {$\because$ P = 1 J/s and m = 2 kg}
Integrating both sides,
$\int\limits_0^3 {2{v^2}dv = \int\limits_0^s {ds \Rightarrow {2 \over 3}[{v^3}]_0^3 = 8} } $
${2 \over 3}[27 - 0] = s \Rightarrow s = 18$ m
Hence, after 9 s, the body has moved a distance of 18 m.
Explanation:
a = g sin 30 + $\mu $ g cos 30
We know, v2 = u2 + 2as
$ \Rightarrow $ 0 = $v_0^2$ - 2ad
$ \Rightarrow $ $v_0^2 = 2ad$
$d = {{v_0^2} \over {2a}}$
Total work done,
${W_f} = {k_f} - {k_i}$
$ \Rightarrow $ $ - 2\mu mg\,\cos 30{{v_0^2} \over {2a}} = {1 \over 2}m{{v_0^2} \over 4} - {1 \over 2}mv_0^2$
$ \Rightarrow $ ${{ + \mu g\,\cos 30} \over a} = $${3 \over 8}$
$ \Rightarrow $ $8\mu g\,\cos 30 = 3g\,\sin 30 + 3\mu \,\cos 30$
$ \Rightarrow $ $5\mu g\,\cos 30 = 3g\,\sin 30$
$ \Rightarrow $ $\mu = {{3\tan 30} \over 5} = {{\sqrt 3 } \over 5}$
$ \Rightarrow $ ${{\sqrt 3 } \over 5} = {I \over {1000}}$
$ \Rightarrow $ $I = 346$
Explanation:
= $\sqrt {2 \times 10 \times 20} $
= 20 m/s
Now work done by the machine,
WF = $\Delta $k
$ \Rightarrow $ F.d = $\Delta $k
$ \Rightarrow $ F = ${{\Delta k} \over d}$
= ${{{1 \over 2} \times 0.15 \times 400 - 0} \over {0.2}}$
= 150 N
A small block starts slipping down from a point B on an inclined plane AB, which is making an angle $\theta $ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction $\mu $. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by $\mu $ = ktan $\theta $ . The value of k is _________.
Explanation:
Apply work energy theorem
Wg + Wf = $\Delta $kE
mgsin$\theta $ (AC + 2AC) – $\mu $mg cos$\theta $ $ \times $ AC = 0 - 0
$ \Rightarrow $ $\mu $ = 3tan $\theta $
(Figure drawn is schematic and not to scale; take g = 10 ms-2)
Explanation:
= 1 × 10 × 2 – 1 × 10 × 1 = 10 J
Explanation:
d = $\alpha $ydx + 2$\alpha $xdy
A$ \to $B, y = 1, dy = 0
then ${W_{A \to B}} = \int {\alpha ydx} $$ = \alpha 1\int_0^1 {dx} = \alpha $
B$ \to $C, x = 1, dx = 0
then ${W_{B \to C}} = 2\alpha 1\int_1^{0.5} {dy} = - 2\alpha (0.5) = - \alpha $
C$ \to $D, y = 0.5, dy = 0
then ${W_{C \to D}} = \int_1^{0.5} {\alpha ydx} = \alpha .{1 \over 2}\int_1^{0.5} {dx} = - {\alpha \over 4}$
D$ \to $E, x = 0.5, dx = 0
then ${W_{D \to E}} = 2\alpha \int {xdy} = 2\alpha .{1 \over 2}\int\limits_{0.5}^0 {dy} = - {\alpha \over 2}$
E$ \to $F, y = 0, dy = 0 then WEF = 0
F$ \to $A, x = 0, dx = 0 then WF$ \to $A = 0
$ \therefore $ $W = \alpha - \alpha - {\alpha \over 4} - {\alpha \over 2} = - {{3\alpha } \over 4}$
Given, $\alpha = - 1 $
$\Rightarrow W = {3 \over 4}J = 0.75J$
Explanation:
After collision the 2.0 kg block will perform simple harmonic oscillation with time period
$T = 2\pi \sqrt {{m \over k}} $
Given m = 2.0 kg and k = 2.0 N m$-$1, we have
T = 2$\pi$
Thus, the block returns to its original position in time
$t = {T \over 2}$ = $\pi$ s
That is, t = 3.14 s
Now, if v1 and v2 are velocities of 1.0 kg block and 2.0 kg block, respectively, before collision; v'1 and v'2 are velocities of 1.0 kg block and 2.0 block, respectively, after collision. So, by conservation of momentum
m1v1 + m2v2 = m1v'1 + m2v'2
Here, m1 = 1.0 kg, v1 is initial speed of 1.0 kg block = 2.0 m s$-$1, m2 = 2.0 kg, v2 is initial speed of 2.0 kg block = 0.0 m s$-$1, v'1 is final speed of 1.0 kg block after collision and v'2 is final speed of 2.0 kg block after collision. Then, 1.0 kg $\times$ 2.0 m/s + 2.0 kg $\times$ 0 m/s = 1.0 v'1 + 2.0 v'2
v'1 + 2v'2 = 2 ..... (1)
Also, using definition of coefficient of restitution
v'2 $-$ v'1 = $\varepsilon $(v1 $-$ v2)
Since collision is elastic, So $\varepsilon $ = 1
$\Rightarrow$ v'2 $-$ v'1 = v1 $-$ v2
$\Rightarrow$ v'2 $-$ v'1 = 2 $-$ 0
$\Rightarrow$ v'2 $-$ v'1 = 2 ...... (2)
From Eqs. (1) and (2), we get
$v{'_2} = {4 \over 3}$ m s$-$1 and $v{'_1} = {-2 \over 3}$ m s$-$1
Therefore, distance between the blocks is given as
$s = v{'_1} \times t = {{ - 2} \over 3} \times 3.14 = 2.09$ m
Explanation:
Work done by the gravitational force is ${W_g} = mgh\cos 180^\circ $
= $-$ mgh = $-$1 $\times$ 10 $\times$ 4 = $-$40 J
Work done by the applied force F
${W_F} = Fd\cos 0^\circ = Fd = 18 \times 5 = 90$ J
According to work-energy theorem
$\Delta$K = Wg + WF
$\Delta$K = $-$40 J + 90 J = 50 J = (5 $\times$ 10) J
$\therefore$ n = 5
Explanation:
To determine the minimum velocity required for the bob of mass $ m $ to complete a full circle in the vertical plane, we start with the condition at the bottommost point:
$ u = \sqrt{5 g l_1} \text{ (at point A)} $
At the highest point B, using the conservation of energy, we get:
$ \frac{1}{2} m u_1^2 = m g (2 l_1) + \frac{1}{2} m v_1^2 $
Simplifying, we have:
$ v_1^2 = u_1^2 - 4 g l_1 $
Substituting $ u_1 = \sqrt{5 g l_1} $:
$ v_1^2 = 5 g l_1 - 4 g l_1 $
$ v_1 = \sqrt{g l_1} $
When the bob at point B elastically collides with another identical mass that is initially at rest, the velocities swap due to the nature of elastic collisions. Thus, the second bob attains the velocity:
$ u_2 = v_1 = \sqrt{g l_1} $
For the second bob to complete its circular path, the minimum velocity requirement is:
$ u_2 = \sqrt{5 g l_2} = \sqrt{g l_1} $
$\therefore \frac{l_1}{l_2}=5$Explanation:
To determine the speed of a particle after $ t = 5 $ seconds, given a constant power, the following steps should be taken :
Power Equation: Power is the rate at which work is done, and work is the change in kinetic energy :
$ P = \frac{d(W)}{dt} = \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) $
Given Data :
Mass of the particle ($ m $) = 0.2 kg
Constant power ($ P $) = 0.5 W
Initial speed ($ v_0 $) = 0 m/s
Differentiating Kinetic Energy with Respect to Time :
$ P = \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) = m v \frac{dv}{dt} $
Rearranging to Solve for $ v \frac{dv}{dt} $ :
$ 0.5 = 0.2 \cdot v \cdot \frac{dv}{dt} $
$ \frac{dv}{dt} = \frac{5}{2v} $
Integrating Both Sides :
$ v \, dv = \frac{5}{2} \, dt $
$ \int v \, dv = \int \frac{5}{2} \, dt $
$ \frac{v^2}{2} = \frac{5t}{2} + C $
Applying Initial Condition (when $ t = 0 $, $ v = 0 $) :
$ 0 = \frac{5 \cdot 0}{2} + C $
$ C = 0 $
Solving for $ v $ :
$ \frac{v^2}{2} = \frac{5t}{2} $
$ v^2 = 5t $
$ v = \sqrt{5t} $
Speed After 5 Seconds :
$ v = \sqrt{5 \cdot 5} = \sqrt{25} = 5 \, \text{m/s} $
Thus, the speed of the particle after 5 seconds is 5 m/s.
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is V = N/10. Then N is
Explanation:
Loss of kinetic energy $ = {1 \over 2}m{V^2}$
Work done against friction $ = \mu mgx$
Gain in potential energy $ = {1 \over 2}k{x^2}$
From work-energy principle,
${1 \over 2}m{V^2} = \mu mgx + {1 \over 2}k{x^2}$
$ \Rightarrow {1 \over 2} \times 0.18 \times {V^2} = 0.1 \times 0.18 \times 10 \times 0.06 + {1 \over 2} \times 2 \times {(0.06)^2}$
$ \Rightarrow V = 0.4 = {4 \over {10}}$ ms$-$1. Hence, N = 4.
A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.
Explanation:
We have,
$a = \left( {{{{m_1} - {m_2}} \over {{m_1} + {m_2}}}} \right)g = \left( {{{0.72 - 0.36} \over {0.72 - 0.36}}} \right) \times 10 = {g \over 3} = {{10} \over 3}$
$T = {{2{m_1}{m_2}g} \over {{m_1} + {m_2}}} = {{2 \times 0.72 \times 0.36 \times 10} \over {0.72 + 0.36}} = 4.8$ N
$s = {1 \over 2}a{t^2} = {1 \over 2} \times {{10} \over 3} \times {1^2} = {5 \over 3}$ m
The work done by the rope on 0.36 kg is
$W = Ts\cos 0^\circ = 4.8 \times {5 \over 3} = + 8$ J
Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m and m, respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of the object C.
Explanation:
Let $V_1$ and $V_2$ be the velocities of blocks A and B immediately after the elastic collision:
${v_1} = \left( {{{{m_1} - {m_2}} \over {{m_1} + {m_2}}}} \right)m = \left( {{{m - 2m} \over {m + 2m}}} \right) \times 9 = - 3$ m/s
${v_2} = \left( {{{2{m_1}} \over {{m_1} + {m_2}}}} \right)m = \left( {{{2m} \over {m + 2m}}} \right) \times 9 = 6$ m/s
After the perfectly inelastic collision between blocks B and C, let $v$ be the common velocity. Applying centre of mass concept, we get
$2m{v_2} = (2m + m)v$
$ \Rightarrow v = {2 \over 3} \times 6 = 4$ m/s
There is a rectangular plate of mass M kg of dimensions ( $a \times b$ ). The plate is held in horizontal position by striking $n$ small balls each of mass m per unit area per unit time. These are striking in the shaded half region of the plate. The balls are colliding elastically with velocity $v$. What is $v$ ?
It is given $n=100, \mathrm{M}=3 \mathrm{~kg}, m=0.01 \mathrm{~kg}$; $b=2 m ; a=1 \mathrm{~m} ; g=10 \mathrm{~m} / \mathrm{s}^2$
Explanation:
The ball collides elastically with the stationary plate. In this collision, velocity of the ball gets reversed.
The conversation of linear momentum,
$ \begin{aligned} \mathrm{P}_{i(\text { ball })}+\mathrm{P}_{i(\text { plate })} & =\mathrm{P}_{f(\text { ball })}+\mathrm{P}_{f(\text { plate })} \\ \Delta \mathrm{P}_{\text {plate }} & =-\Delta \mathrm{P}_{\text {ball }} \\ & =-m v-(-m v) \\ & =2 m v \text { (upwards). } \end{aligned} $
number of balls striking the plate in times $\Delta t$ is $\mathrm{N}=n(a b / 2) \Delta \mathrm{t}=n a b \Delta t / 2$.
Thus, the total change in the plate's momentum in time $\Delta t$ is
$ \Delta \mathrm{P}=\mathrm{N} \Delta \mathrm{P}_{(\text {plate })}=\text { mvnab } \Delta t . $
By Newton's second law, upward force on the
$ \text { plate is } \mathrm{F}=\frac{\Delta \mathrm{P}}{\Delta t}=m v n a b .\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $
This force effectively acts at the centre of the shaded region.
i.e., at a distance $\frac{3 b}{4}$ from the hinge.
In equillibrium $\mathrm{F}+\mathrm{R}=m g$, and torque due to all the forces about any point is zero.
The torque about a point on the hinge is
$ \mathrm{F}\left(\frac{3 b}{4}\right)-\operatorname{Mg}\left(\frac{b}{2}\right)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$
Substitute value of F from equation (i) into equation (ii) to get
$ \begin{aligned} \mathrm{F} & =\frac{\mathrm{Mg} b}{2} \times \frac{4}{3 b}=\frac{\mathrm{Mg} 2}{3}=\text { mvnab } \\ v & =\frac{q 2 \mathrm{Mg}}{3 m n a b}=\frac{2 \times 3 \times 10}{3 \times 0.01 \times 100 \times 1 \times 2} \\ & =\frac{60}{6}=10 \mathrm{~m} / \mathrm{s} \end{aligned} $