Work Energy and Power Test

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Work Energy and Power Test - 31

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Question 1
1 / -0

A force acts on a $$3$$ gm particle in such a way that the position of the particle as a function of time is given by $$x=3t-4t^2+t^3$$, where x is in meters and t is in seconds. The work done during the first $$4$$ second is:

A
$$384 mJ$$

B
$$168 mJ$$

C
$$528 mJ$$

D
$$541 mJ$$

Solution

Mass of the particle,
$$m=0.003kg$$
$$x=3t-4t^2+t^3$$
$$\therefore a=-8+6t$$
Thus force acting on the particle is $$ma$$
Thus the work done on the particle$$=\int Fdx$$
$$=\int_0^4 m(-8+6t)(3-8t+3t^2)dt$$
$$=528mJ$$ -m represents mass here.
Question 2
1 / -0

The work done by the frictional force on a surface in drawing a circle of radius r on the surface by a pencil of negligible mass with a normal pressing force N (coefficient of friction $$\mu_k$$ ) is :

A
$$4\pi r^{2} \mu_K N$$

B
$$-2\pi r^{2} \mu_K N$$

C
$$-2\pi r \mu_K N$$

D
Zero

Solution

The mass of the pencil is negligible, and the work done by normal pressing force causes the pencil tip to reduce. Hence, there is no work done by frictional force.
Question 3
1 / -0

A bead is free to slide down a smooth wire tightly stretched between points $$A$$ and $$B$$ on a verticle circle of radius $$R$$. If the bead starts from rest at '$$A$$', the highest point on the circle, its velocity when it arrives at $$B$$ is

A
$$2\sqrt{gR}$$

B
$$2\sqrt{gR} \cos{\theta}$$

C
$$\displaystyle \dfrac{2\sqrt{R}}{g}$$

D
None of the above

Solution

$$a= g\cos{\theta}\\AB = 2R\cos{\theta}\Rightarrow v^2 =u^2 + 2as \\v^2 = 0 + 2 (g\cos{\theta}) 2R\cos{\theta}\Rightarrow v^2 = 4gR \cos ^{2 }{\theta}\\ \Rightarrow v = 2\sqrt{ gR} \cos{\theta}$$
Question 4
1 / -0

A bob hangs from a rigid support by an inextensible string of length $$\ell$$. If it is displaced through a distance $$\ell$$ (from the lowest position) keeping the string straight, & then released. The speed of the bob at the lowest position is:

A
$$\sqrt{g\ell}$$

B
$$\sqrt{3g\ell}$$

C
$$\sqrt{2g\ell}$$

D
$$\sqrt{5g\ell}$$

Solution

Distance= $$l$$
change in height= $$lcos60 = l/2$$

From energy conservation,
PE at max height= KE at lowest point= $$mgl/2= mv^2/2$$
$$v=\sqrt{gl}$$
Question 5
1 / -0

A ball of mass 3 kg collides with a wall with velocity $$10\:m/sec$$ at an angle of $$30^{\circ}$$ with the wall and after collision reflects at the same angle with the same speed. The change in momentum of ball in MKS unit is-

A
20

B
30

C
15

D
45

Solution

$$\Delta p=2mv\:\cos 60^{\circ}=mv= 2 \times 3 \times 10 \times \frac{1}{2}=30\: kg \:m/s$$
Question 6
1 / -0

Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass during the pulling is

A
$$\displaystyle \frac{1}{2}kx^{2}$$

B
$$\displaystyle -\frac{1}{2}kx^{2}$$

C
$$\displaystyle \frac{1}{4}kx^{2}$$

D
$$\displaystyle -\frac{1}{4}kx^{2}$$

Solution

Hint:
Whenever we extend or compress a spring of spring constant$$k$$, spring exerts a force on us that opposes the change in length of spring.

Step 1: Calculate the initial and Final Energy of the system.
Let the mass of two identical blocks attached to two ends of spring is $$m$$. Let spring constant of spring is $$k$$, and length of spring is $$L$$.
When the spring is in its original length, the Energy stored in the system is, $$E_1 = 0$$.
When spring is stretched by length x, by pulling both the blocks symmetrically, Energy stored in the system is given by,
$$E_2 = \dfrac{1}{2}kx^2$$

Step 2: Calculate Work done by spring.
By conservation of Potential Energy, we know that work done by a conservative force is equal to change in Potential Energy of the system.
Therefore, work done by spring is,
$$ W = -(\dfrac{1}{2}kx^2 - 0)$$

$$\Rightarrow W = -\dfrac{1}{2}kx^2$$
Since, both the blocks are pulled symmetrically, work done by spring on each block will behalf of this work done.
Therefore Work done by spring on each block is,
$$w = -\dfrac{1}{4}kx^2$$

Thus, Work done by spring on each mass is, $$w = -\dfrac{1}{4}kx^2$$
Option D is correct.
Question 7
1 / -0

What is work done in holding a body of mass $$20\ kg$$ at a height of $$2\ m$$ above the ground? $$(g = 10\ m/s^2)$$

A
$$40\ J$$

B
$$400\ J$$

C
$$10\ J$$

D
Zero

Solution

A body of mass $$20\ kg$$ is held at a height of $$2\ m$$ above the ground means there is no displacement because of which there is no change in its potential energy. Hence work done is zero.
Question 8
1 / -0

The velocity of a bus, moving on a smooth road, is increased from 8 m/s to 32 m/s in 120s. During this process, the potential energy of the bus

A
Does not change

B
Becomes twice that of initial potential energy

C
Becomes four times that of initial potential energy

D
Becomes sixteen times that of initial potential energy

Solution

As potential energy, $$P=mgh$$
Since, there is no vertical displacement so, $$h=0$$.
Hence, potential energy change$$P=mgh=0$$
The moving bus has only change in its Kinetic energy.
Question 9
1 / -0

Two bodies P and Q of equal masses are kept at heights $$x$$ and $$4x$$ respectively. What will be the ratio of their potential energies?

A
$$1:8$$

B
$$4:1$$

C
$$1:4$$

D
$$8:1$$

Solution

Potential energy $$P=mgh$$
Given, $$h_1=x ; h_2=4x$$

Since, the masses are same,
then $$\dfrac{P_1}{P_2}=\dfrac{h_1}{h_2} = \dfrac{x}{4x}=1:4$$
Question 10
1 / -0

Initially spring is relaxed. A person starts pulling the spring by applying a variable force $$F$$. Find out the work done by $$F$$ to stretch it slowly to a distance by $$x$$.

A
$$\dfrac{Kx^2}{1}$$

B
$$\dfrac{Kx^2}{2}$$

C
$$\dfrac{Kx^2}{4}$$

D
$$\dfrac{Kx^2}{8}$$

Solution

$$\int dW\, =\, \int F\, . \, ds\, =\, \int_0^x Kxdx

\quad\, \Rightarrow \quad\, W\, =\,\displaystyle { \left (

\frac{kx^2}{2} \right )^x_{\circ}\, =\, \frac{Kx^2}{2}}$$