Work Power & Energy
A stone is projected at angle $30^{\circ}$ to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be -
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 ball is projected with kinetic energy E, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :
A bullet of mass $200 \mathrm{~g}$ having initial kinetic energy $90 \mathrm{~J}$ is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to $40 \mathrm{~J}$ within $1 \mathrm{~s}$, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is
Sand is being dropped from a stationary dropper at a rate of $0.5 \,\mathrm{kgs}^{-1}$ on a conveyor belt moving with a velocity of $5 \mathrm{~ms}^{-1}$. The power needed to keep the belt moving with the same velocity will be :
As per the given figure, two blocks each of mass $250 \mathrm{~g}$ are connected to a spring of spring constant $2 \,\mathrm{Nm}^{-1}$. If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is :
A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms$-$1 gets embedded in it, then loss of kinetic energy will be :
A body of mass $0.5 \mathrm{~kg}$ travels on straight line path with velocity $v=\left(3 x^{2}+4\right) \mathrm{m} / \mathrm{s}$. The net workdone by the force during its displacement from $x=0$ to $x=2 \mathrm{~m}$ is :
In the given figure, the block of mass m is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is
A particle of mass 500 gm is moving in a straight line with velocity v = b x5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m$-$3/2 s$-$1).
Arrange the four graphs in descending order of total work done; where W1, W2, W3 and W4 are the work done corresponding to figure a, b, c and d respectively.
A particle experiences a variable force $\overrightarrow F = \left( {4x\widehat i + 3{y^2}\widehat j} \right)$ in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then Kinetic Energy changes by :
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
[ where K.E. = kinetic energy ]
(take g = 9.8 ms$-$2)
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