Work Energy and Power Test

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Work Energy and Power Test - 17

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Question 1
1 / -0

A force of $$16N$$ is distributed uniformly on one surface of a cube of edge $$8cm$$. The pressure on this surface is

A
$$3500Pa$$

B
$$2500Pa$$

C
$$4500Pa$$

D
$$5500Pa$$

Solution

$$F=16N$$
$$A=8\times 8\times { 10 }^{ -4 }{ m }^{ 2 }$$
$$P=\cfrac { F }{ A } =\cfrac { 16 }{ 64\times { 10 }^{ -4 } } =\cfrac { 1000 }{ 4 } =2500Pa$$
Question 2
1 / -0

In the phenomenon of work done by variable forces, the forces:

A
remain constant

B
don't remain constant

C
increase

D
decrease

Solution

The variable forces are the non-constant forces that changes with maybe time,distance or any other variable.
hence option B is correct that states forces don't remains constant
Question 3
1 / -0

A brick of mass $$m$$ , tied to a rope , is being whirled in a vertical circle, with a uniform speed. The tension in the rope is:

A
The same throughout

B
Largest when the brick is at the highest point of the circular path and smallest when it is at the lowest point

C
Largest when the rope is horizontal and smaller when it is vertical.

D
Largest when the brick is at the lowest point of the circular path and smallest when it is at the highest point

Solution

$$ T_H + mg = m \omega^2 r$$
$$ T_L - mg = m\omega^2 r$$
$$ \therefore T_L > T_H$$
Question 4
1 / -0

A $$\vec{F}=(5\hat{i}+3\hat{j}+2\hat{k})\ N$$ is applied over a particle with displaces it from its origin to the point $$\vec{r}=(2\hat{i}-\hat{j})m$$. The work done on the particle is joule is

A
$$-7\ J$$

B
$$+7\ J$$

C
$$+10\ J$$

D
$$13\ J$$

Solution

Work done by a force =
$$\int \vec{F}.\vec{dr}=(5\hat{i}+3\hat{j}+2\hat{k}).(2\hat{i}-\hat{j})=7 \ Joule$$
Question 5
1 / -0

A particle moves from $$X = 0\ $$ to $$X = 2\ m$$ on X-axis under the effect of $$ { F } (x) = ({ 4x }^{ 3 } - { 3x }^{ 2 } + 2x + 5)\widehat { i } $$ Newton. The work done on the particle is

A
$$ 22\ Joule$$

B
$$ 36\ Joule$$

C
$$ 46\ Joule$$

D
None of these

Solution

Work done by a force =
$$\int Fdx=x^4-x^3+x^2+5x|^2_0=22 \ Joule$$
Question 6
1 / -0

A force F = $$-\frac{K}{X^2} (X \neq 0)$$ acts on a particle in x-direction. The work done by this force in displacing the particle from x = +a to x = +2a is. (Where k is a positive constant)

A
$$\frac{-k}{2a}$$

B
$$\frac{+k}{2a}$$

C
$$\frac{k}{a}$$

D
$$-\frac{k}{a}$$

Solution

Given that , $$F= - \dfrac K{x^2}$$ , Where $$x$$ is the position of particle .

Work done by this force , $$W=\int _a^{2a} F.dx=\int _a^{2a} -\dfrac K{x^2} dx= [\dfrac Kx ]_a ^{2a}= \dfrac K{2a}- \dfrac Ka = -\dfrac K {2a} $$
Question 7
1 / -0

Given $$k_1 = 1500 \ N/m,$$ $$k_2 =500 \ N/m,$$ $$m_1 = 2 \ kg$$ and $$m_2 = 1 \ kg$$. Find potential energy stored in equilibrium: (Take $$g = 10 \ m/s^2$$)

A
2.3 J

B
0.4 J

C
3.4 J

D
0.6 J

Solution

Soppose the elongation in first string be $$x_1$$ and in the second string be $$x_2$$
Now for second string, in equilibrium
$$K_2x_2=m_2g$$
or $$x_2= m_2g/k_2$$ (1)
For first string in equilibrium
$$K_1x_1=(m_1+m_2)g$$
or $$x_1= (m_2+m_1)g/k_1$$ (2)
Now potential energy stored will be
$$U=\frac{K_1x_1^2}{2}$$ + $$\frac{K_2x_2^2}{2}$$
Putting the values of $$k_1$$,$$k_2$$,$$x_1$$ and $$x_2$$
We get U= 0.4J
Question 8
1 / -0

A ball of mass $$'m'$$ moves towards a wall with a velocity $$'u'$$, the direction of motion making an angle $$\theta$$ with the surface of the wall and rebounds with the same period. The change in momentum of the ball during the collision is:

A
$$2mu\ sin$$$$\theta$$ towards the wall

B
$$2mu \ sin$$$$\theta$$ away from the wall

C
$$2mu\ cos$$$$\theta$$ towards the wall

D
$$2mu\ cos$$$$\theta$$ away from the wall

Solution

As the ball hits the wall at an angle and rebounds, there is a change in the momentum only along the direction perpendicular to the wall. The change along the wall is zero. In diagram $$\vec p_1$$ and $$\vec p_2$$ is the momentum before and after the collision with wall.
Since , the angle with the wall is $$\theta $$, the change is only in $$sin\theta $$ component, which is across $$\vec F_{wall}$$.
Change in momentum$$=m[usin\theta -(-usin\theta )]=2mu\ sin\theta $$ away from the wall, as the final velocity is away from the wall.
Question 9
1 / -0

A car of mass 400 kg travelling at 72 kmph crashes a truck of mass 4000 kg and travelling at 9 kmph in the same direction. The car bounces back with a speed of 18 kmph. The speed of the truck after the impact is

A
9 kmph

B
18 kmph

C
27 kmph

D
36 kmph

Solution

Using momentum conservation,

$$\displaystyle m_{car} \displaystyle u _{car} + \displaystyle m_{ truck} \displaystyle u_{ truck}= \displaystyle m_{ car} \displaystyle \displaystyle v _{car} + \displaystyle m_{truck} \displaystyle v_{truck}$$

$$400 \times 72 + 4000 \times 9 = -18 \times 400 + 4000 \times v$$

$$v = 18 \ kmph$$
Question 10
1 / -0

A perfectly elastic ball $$p_{1}$$of mass 'm' moving with velocity v collides elastically with three exactly similar balls $$p_{2}$$ , $$p_{3}$$ , $$p_{4}$$ lying on a smooth table. Velocities of the four balls after collision are

A
0,0,0,0

B
v,v,v,v

C
v,v,v,0

D
0,0,0,v

Solution

$$\textbf{Hint}$$: Use law of conservation of momentum
$$\textbf{Step}$$:
Here when $$P_{1}$$ and $$P_{2}$$ will collide, their momentum will be exchanged because the collision is elastic and the two bodies are similar.
When $$P_{2}$$ and $$P_{3}$$ collide same happens.
So, momentum of $$P_{1}$$ is transferred to $$P_{4}$$
So, $$P_{1}$$, $$P_{2}$$, $$P_{3}$$ will have zero velocity and $$P_{4}$$ has a velocity of $$V$$
$$\textbf{Step 2}: \textbf{Calculation}$$:
Let us take two balls. Let the first ball has a mass of 'm' and velocity 'v'. Let the second ball has a mass of 'm' and velocity is $$0$$ that is it is at rest.
For perfectly elastic collision coefficient of restitution is $$e=1$$
Total momentum before collision (Initial momentum)=Total momentum after collision(Final momentum)
$$P_i=P_f$$
$$mV=mV_1+mV_2$$
$$\implies V_1+V_2=V$$ $$(1)$$
We know coefficient of restitution $$e=1$$
So, $$e= \dfrac{V_{separation}}{V_{approach}}$$
$$1=\dfrac{V_1-V_2}{V}$$
$$V_1-V_2=V$$ $$(2)$$
Solving equation $$(1)$$ and $$(2)$$ we get
$$2V_1=2V$$
$$\implies V_1=V$$
$$V_2=0$$
So $$P_{1}$$, $$P_{2}$$, $$P_{3}$$ will have zero velocity and $$P_{4}$$ has a velocity of $$V$$
Thus option D is correct.