Work Energy and Power Test

Result

Work Energy and Power Test - 50

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A $$10$$ kkg object attached to a nylon cord outside a space vehicle is rotating at a speed of $$5 m/s$$. If the force acting on the cord is $$125 N$$ its radius of path is

A
2 m

B
4 m

C
6 m

D
1 m

Solution
Question 2
1 / -0

A ball falls from a height such that it strikes the floor of lift at $$10\ m/s$$, if lift is moving in the upward direction with a velocity $$1\ m/s$$ , then velocity with the ball rebounds after elastic collision will be then

A
$$11\ m/s$$

B
$$12\ m/s$$

C
$$13\ m/s$$

D
$$9\ m/s$$

Solution

Assume upwards is positive, the lift (object 1) is moving up with speed $$1 m/s$$.
$$e = \dfrac {v_2 - v_1}{u_1 - u_2} $$ and collision is elastic.
$$u_1 = 1m/s$$, $$u_2 = -10m/s$$
$$v_1 = 1m/s$$
$$ 1 = \dfrac {v_2 - 1}{1 - (-10)} $$
$$v_2 = 11\ m/s$$
Question 3
1 / -0

The length of simple pendulum is 1 m and mass of its bob is 50 g. The bob is given sufficient velocity so that the bob describe vertical circle whose radius equal to length of pendulum. The maximum difference in the kinetic energy of bob during one revolution is.

A
0.98 J

B
1.96 J

C
4.9 J

D
9.8 J

Solution

A bob of mass $$m = 0.05\ Kg$$ is performing vertical circle of radius $$(r) = 1m$$.
In the boundary case.
a minimum velocity of $$\mu = \sqrt{5gr}$$ has to provided to it
so it completes a vertical circle.
So, kinetic energy $$(K_1)$$ at the lowest point,
$$K_1 = \dfrac{1}{2}m\mu^2 = \dfrac{5}{2}mgr$$
and $$PE = 0$$, raking it as reference at highest point
$$PE : \mu_2 =mgh = 2mgr$$ $$(as \, h = 2r)$$
From conservation of energy.
$$K_1 + \mu_1= K_2+\mu_2 \Rightarrow K_2 = K_1 - \mu_2 = \dfrac{1}{2}mgr$$
Thus, loss in $$KE$$:
$$\Delta K = K_1-K_2 = 2mgr = 2\times 0.05 \times 9.8 \times 1 = 0.98\ J$$ (Ans)
Question 4
1 / -0

The work done by a force $$\vec{F}=\left ( -6x^3\hat{i} \right )\ N$$ is displacing a particle from $$x=4\ m$$ to $$x=-2\ m$$ is

A
$$-240\ J$$

B
$$360\ J$$

C
$$120\ J$$

D
$$420\ J$$

Solution

$$F=-6{ x }^{ 3 }\hat { i } $$
$$x=4$$ to $$x=-2$$
$$\int { dw } =\int { f.dx } $$
As displacement is along $$\hat { i } $$ so $$F=-6{ x }^{ 3 }$$
$$-\int { _{ 4 }^{ -2 } } 6{ x }^{ 3 }dx=\left[ \cfrac { 6{ x }^{ 4 } }{ 4 } \right] ^{ -2 }_{ 4 }$$
$$\left[ \cfrac { -6{ x }^{ 4 } }{ 4 } +\cfrac { 6{ x }^{ 4 } }{ 4 } \right] _{ 4 }^{ -2 }$$
$$\cfrac { -6\times { (4) }^{ 4 } }{ 4 } +\cfrac { 60{ (-2) }^{ 4 } }{ 4 } $$
$$=360J$$
Question 5
1 / -0

When a spring is stretched by a distance $$x,$$ it exerts a force given by $$F$$ $$=$$ ($$ -5x - 16x^3$$) $$N,$$ where $$x$$ is in $$m.$$ The work done, when the spring is stretched from $$0.1\ m$$ to $$0.2\ m$$ is :

A
$$6.9\times 10^{-2}\ J$$

B
$$12.2 \times 10^{-2}\ J$$

C
$$8.1 \times 10^{-2}\ J$$

D
$$12.2 \times 10^{-1}\ J$$

Solution

Given,

$$F=-5x-16x^3$$

$$-kx=-5x-16x^3$$

$$k=5+16x^2$$

work done,

$$=\dfrac{1}{2}(k_1x_1^2+k_2x_2^2)$$

upon solving the above equation, we get,

work done $$=8.7 \times 10^{-2}J$$
Question 6
1 / -0

A body of mass tied at the end of a strong of length $$l$$ is projected with velocity $$\sqrt{4lg}$$, at what height will it leave the circular path:

A
$$\dfrac{5}{3}l$$

B
$$\dfrac{3}{5}l$$

C
$$\dfrac{1}{3}l$$

D
$$\dfrac{2}{3}l$$

Solution

REF.Image
In a vertical circle the tension at the string is given by :-
$$T= \dfrac{mu^{2}}{l}-mg(2-3 cos \theta )$$
Here $$u^{2}= 4gl$$, and particle stop circular motion
when $$T=0$$
$$\Rightarrow T=0\Rightarrow \dfrac{mu^{2}}{L}= mg(2-3 cos\theta )$$
$$\Rightarrow \dfrac{m.4gL}{L}= mg(2-3 cos \theta )\Rightarrow cos\theta = -2/3$$
$$\therefore $$ Height, $$h=l\left | cos\theta \right |=\frac{2}{3}l$$
Hence it attains maximum height,
$$H= h+L= \dfrac{5}{3}L$$ (Ans)
Question 7
1 / -0

A ball of mass m approaches a wall of mass M (>> m) with the speed 4 m/s along normal to the wall. The speed of wall is 1m/s towards the ball . The speed of the ball after an elastic collision with the wall is-

A
5 m/s away from the wall

B
3 m/s away from the wall

C
9 m/s away from the wall

D
6 m/s away from the wall

Solution

$$\textbf{Step 1: Initial & final situation [Ref. Fig.]}$$
Right direction is positive and left direction is negative.

$$\textbf{Step 2: Apply law of restitution}$$
Final velocity of wall be same as initial as $$M>>> m$$
Therefore $$V_{wall} = -1m/s$$

$$e = \dfrac{V(seperation)}{V(approach)} $$ (Along line of collision)

For elastic collision $$e=1$$

$$\Rightarrow e = 1 = \left(\dfrac{V_{wall}-V_{ball} }{u_{ball} - u_{wall}} \right)$$

$$u_{wall} =- 1m/s$$
$$u_{ball} = +4 \, m/s$$

$$\Rightarrow 1 = \left(\dfrac{-1-V_{ball} }{4 + 1} \right) \Rightarrow V_{ball} = -6 \,m/s$$
Negative sign means left direction (away from wall)
Hence option D is correct.
Question 8
1 / -0

Find the reading of machine. Given each box of mass$$=15 kg$$,mass of man $$ 1=30kg$$, mass of man $$2=40 kg$$ and of each weighing machine $$=5 kg$$

A
$$360/11 kg, 400/11 kg $$

B
$$330/11 kg, 440/11 kg $$

C
$$360/11 kg, 440/11 kg $$

D
$$350/11 kg, 240/11 kg $$

Solution

$$Given:$$ Each box of mass$$=15kg$$ ; mass of man$$1 = 30kg$$ ; mass of man$$ 2 = 40kg$$ ; each Weighing machine$$=5kg$$
$$Solution:$$
$$T-50 g=50a ....(i) $$
$$60g -T=60a ....(ii)$$
$$\Rightarrow 10g =110a$$
$$\Rightarrow a=\dfrac{g}{11}$$
$$For (i) $$
$$Reading =m g+m a$$
$$=30 g+\dfrac{30}{11}g$$
$$=\dfrac{360}{11} g$$.
In $$kg=\dfrac{360}{11}kg$$
$$For(ii)$$
$$\Rightarrow mg-ma$$
$$=\dfrac{400}{11} k g$$ .
$$So,the$$ $$correct$$ $$option:A$$
Question 9
1 / -0

When a massive body suffers an elastic collision with a stationary light body, then massive body approximately comes to rest and light body-

A
acquires velocity greater than initial velocity of massive body

B
sticks to the massive body and remains at rest.

C
acquires half the initial velocity of the massive body

D
remains at rest but does not stick to the massive body.

Solution
Question 10
1 / -0

A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at $$x=0$$ has the kinetic energy of $$25$$ J, then the kinetic energy of the particle at $$x=16$$m is?

A
$$45$$J

B
$$30$$J

C
$$70$$J

D
$$135$$J