Work, Energy and Power

Work done in stretching a string to obtain an extension l,

$ W_1=\frac{1}{2} k l^2 $

Similarly, work done in stretching a string to obtain an extension $l_1$ is,

$ W_2=\frac{1}{2} k l_1^2 $

∴ Work done in second case, $W$

$ =W_2-W_1=\frac{1}{2} k\left(l_1^2-l^2\right) $

$ \begin{aligned} & \text {As, } F=\mu m g \cos \theta \\ & \text { or } \quad F=0.30 \times 10 \times 10 \cos 45^{\circ} \\ & \text { or } \quad F=\frac{30}{\sqrt{2}} \mathrm{~N} \end{aligned} $

$ \begin{aligned} W & =F \times s \times \cos 180^{\circ}=-\frac{30}{\sqrt{2}} \times 5 \\ & =-\frac{150}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=-75 \sqrt{2} \mathrm{~J} \end{aligned} $

As the vertical component of the velocity at highest point is zero.

$ u_y=0, u_x=u \cos \beta $

The kinetic energy at the highest point will be equal to $\frac{1}{2} m(u \cos \beta)^2$

The initial kinetic energy $=\frac{1}{2} m u^2$

$\therefore \quad \frac{1}{2} m(u \cos \beta)^2=\frac{3}{4} \times \frac{1}{2} m u^2$

$\Rightarrow \quad \cos \beta=\frac{\sqrt{3}}{2} \Rightarrow \beta=30^{\circ}$

To determine the work done by the lower block on the upper block, follow these steps:

Calculate Maximum Possible Acceleration:

The maximum acceleration the upper block (1 kg) can experience without sliding off is due to static friction:

$ a_{\text{max}} = \mu g = 1 \, \text{m/s}^2 $

where $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.

Calculate Common Acceleration:

The applied force is distributed over the total mass (3 kg) of the system, so the acceleration $a$ is:

$ a = \frac{F}{m_{\text{total}}} = \frac{0.5}{3} = \frac{0.5}{3} \, \text{m/s}^2 $

Check for Relative Motion:

Since $ a < a_{\text{max}} $, the blocks will move together without slipping.

Calculate the Force of Friction:

The force of friction acting from the lower block to the upper block is:

$ f = m_{\text{upper}} \cdot a = 1 \cdot \frac{0.5}{3} = \frac{1}{6} \, \text{N} $

Calculate the Work Done by Friction:

The work done by this friction force over a displacement of 3 meters is:

$ W = f \times s = \frac{1}{6} \times 3 = 0.5 \, \text{J} $

Thus, the work done by the lower block on the upper block for a displacement of 3 meters is $0.5 \, \text{J}$.

Power of tension $ =0 $

Power of $ m g=(m g)(v) \cos 60^{\circ} $

Here, $ \quad v=\sqrt{2 g h} $

and $ \quad h=l\left(\cos 30^{\circ}-\cos 60^{\circ}\right)=0.36 \mathrm{~m} $

$ v=\sqrt{2 \times 10 \times 0.36} $

$ =2.68 \mathrm{~m} / \mathrm{s} $

Power $ =(m g)(v) \cos 60^{\circ} $

$ \begin{array}{l} =(1 \times 10)(268) \times \frac{1}{2} \\ =1.34 \times 10=13.4 \mathrm{~W} \end{array} $

To find the speed of the mass just before it touches the spring, we begin by determining the spring constant, $ k $, using Hooke's Law, where the force $ F $ is proportional to the compression $ x $:

$ F = kx $

Given that a force of 100 N compresses the spring by 1 m, the spring constant $ k $ is calculated as:

$ k = \frac{F}{x} = \frac{100 \, \text{N}}{1 \, \text{m}} = 100 \, \text{N/m} $

Next, we apply the principle of energy conservation. The initial potential and kinetic energy will equal the final potential energy stored in the compressed spring. The relevant energy conservation equation is:

$ \frac{1}{2} mv^2 + mgh = \frac{1}{2} kx_{\text{max}}^2 $

where:

$ m $ is the mass of the block (10 kg),

$ g $ is the acceleration due to gravity (10 m/s$^2$),

$ h $ is the vertical height change, which is the same as the compression in this scenario ($x_{\text{max}} = 2 \, \text{m}$),

$ v $ is the speed just before touching the spring,

$ x_{\text{max}} $ is the maximum compression of the spring (2 m).

Rearranging the equation for $ v $ gives:

$ v = \sqrt{\frac{k x_{\text{max}}^2}{m} - 2gh} $

Substituting the known values:

$ v = \sqrt{\frac{(100 \, \text{N/m})(2 \, \text{m})^2}{10 \, \text{kg}} - 2(10 \, \text{m/s}^2)\left(\frac{2}{2}\right)} $

This simplifies to:

$ v = \sqrt{\frac{(100)(4)}{10} - 40} $

$ v = \sqrt{40 - 20} = \sqrt{20} \, \text{m/s} $

Thus, the speed of the mass just before it touches the spring is $ \sqrt{20} \, \text{m/s} $.

To find the force acting on a particle in a force field given by a potential energy function, we can use the relationship between force and potential energy. The force is the negative gradient of the potential energy function:

$\mathbf{F} = -\nabla u(x, y)$

Given the potential energy function:

$u(x, y) = \cos(x + y)$

The gradient operator $\nabla$ in two dimensions is given by:

$\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\right)$

To find the force, we first need to compute the partial derivatives of $u(x, y)$ with respect to $x$ and $y$.

Calculate the partial derivative of $u(x, y)$ with respect to $x$:

$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\cos(x + y) = -\sin(x + y)$

Now, calculate the partial derivative of $u(x, y)$ with respect to $y$:

$\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}\cos(x + y) = -\sin(x + y)$

So the gradient of $u(x, y)$ is:

$\nabla u(x, y) = \left(-\sin(x + y), -\sin(x + y)\right)$

The force $\mathbf{F}$ is the negative of this gradient:

$\mathbf{F} = -\nabla u(x, y) = \left(\sin(x + y), \sin(x + y)\right)$

We need to evaluate this at the coordinates $(0, \pi / 6)$:

$ (x, y) = (0, \pi / 6) $

Substitute these values into the sine function:

$ \sin(0 + \pi / 6) = \sin(\pi / 6) = \frac{1}{2} $

Thus, the force $\mathbf{F}$ at $(0, \pi / 6)$ is:

$\mathbf{F} = \left(\frac{1}{2}, \frac{1}{2}\right)$

In vector notation, this can be written as:

$ \mathbf{F} = \frac{1}{2} \hat{\mathbf{i}} + \frac{1}{2} \hat{\mathbf{j}} $

This matches with Option B:

$\frac{1}{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}})$

work done by the gun = total kinetic energy of the bullets

$ = n{1 \over 2}m{v^2} = 300 \times {1 \over 2} \times 10 \times {10^{ - 3}} \times {(600)^2}$

$ = 150 \times 10 \times {10^{ - 3}} \times 600 \times 600$

Power of gun $ = {{work\,done} \over {Time\,taken}}$

$ = {{150 \times 10 \times {{10}^{ - 3}} \times 600 \times 600} \over {60\,s}}$

= 9 kW ($\because$ 1 minute = 60 seconds)

Work done by the particle in X-direction is given by

dW = F dx = (a + bx) dx

Total work done in displacement from x = 0 to x = d will be

$W = \int {dW = \int\limits_0^d {(a + bx)dx} } $

$ = \left[ {ax + {{b{x^2}} \over 2}} \right]_0^d$

$ \Rightarrow W = \left( {a + {{bd} \over 2}} \right)d$

Kinetic energy at Q = 90% of potential energy at P

$ \Rightarrow {1 \over 2}m{v^2} = {{90} \over {100}} \times mgh$

$ \Rightarrow v = \sqrt {1.8\,gh} \Rightarrow v = \sqrt {1.8 \times 10 \times 2} $

$ \Rightarrow v = 6$ m/s

Given, initial speed of ball, v = 16 ms^$-$1

Maximum height, h = 9 m

Mass of ball, m = 0.5 kg

Energy supplied to ball while throwing up,

${E_1} = {1 \over 2}m{v^2} = {1 \over 2} \times 0.5 \times 16 \times 16 = 64$ J

Energy stored when ball reaches the maximum height,

E₂ = mgh = 0.5 $\times$ 9.8 $\times$ 9 = 44.1 J

Energy dissipated by air drag acting on the ball during the ascent,

E = E₁ = E₂ = 64 $-$ 44.1 = 19.9 J

Given, force, F = (8t) N, Mass, m = 2 kg

From Newton's second law,

Force = Rate of change of momentum

$F = {{dp} \over {dt}}$

i.e. dp = Fdt

$\therefore$, $p = \int {dp = \int_0^1 {F\,.\,dt = \int_0^1 {8t\,.\,dt} } } $

$ = 8\left[ {{{{t^2}} \over 2}} \right]_0^1 = 8\left( {{1 \over 2} - 0} \right)$

$\Rightarrow$ p = 4 kg ms^$-$1

Change in kinetic energy, $\Delta K = {{{p^2}} \over {2m}} = {{{{(4)}^2}} \over {2 \times 2}} = 4$ J

From work-energy theorem,

Work done = Change in kinetic energy

So, work done by the force during the first 1s,

W = $\Delta$K = 4J

Work done, W = area under F-s graph

$\therefore$ Total work done is covering an initial distance of 40 m,

W = area of trapezium ABCD + area of trapezium CEFD

$ = {1 \over 2} \times 20(20 + 30) + {1 \over 2} \times 10 \times (20 + 40) = 800$ J

[$\because$ Area of trapezium = ${1 \over 2}$ $\times$ height $\times$ sum of parallel sides]

$\Rightarrow$ W = 800 J