Work Energy and Power Test

Result

Work Energy and Power Test - 74

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 particle of mass m starts moving from origin along x-axis and its velocity varies with position (x) as $$v$$ = $$k\sqrt{x}$$. The work done by force acting on it during first 't' seconds is

A
$$\frac{mk^4t^2}{4}$$

B
$$\frac{mk^2t}{2}$$

C
$$\frac{mk^4t^2}{8}$$

D
$$\frac{mk^2t^2}{4}$$

Solution

$$\textbf{Step 1 - Calculate velocity in terms of t.}$$
Given, $$v = k\sqrt {x}$$

$$v = \dfrac {dx}{dt} = k\sqrt {x}$$

$$kdt = x^{-\dfrac {1}{2}} dx$$

Integrating both side
$$\int kdt = \int x^{-\dfrac {1}{2}} dx$$

$$kt = \dfrac {x^{\dfrac {1}{2}}}{\dfrac {1}{2}} = 2\sqrt {x}$$

$$\Rightarrow x = \dfrac {k^{2}t^{2}}{4}$$

$$\Rightarrow v = k\sqrt {x} = k \sqrt {\dfrac {k^{2}t^{2}}{4}}$$

$$\Rightarrow v = \dfrac {k^{2}t}{2}$$

$$\textbf{Step 2 - By Work-Energy theorem}$$
$$W = \triangle k \Rightarrow W = k_{f} - k_{i}$$

$$\Rightarrow W = \dfrac {1}{2} mv_{f}^{2} - \dfrac {1}{2}mv_{i}^{2}$$

$$= \dfrac {1}{2} m\left (\dfrac {k^{2}t}{2}\right )^{2} - 0 (at\ x_{i} = 0, v_{i} = 0)$$

$$\Rightarrow W = \dfrac {mk^{4} t^{2}}{8}$$

Correct option : C
Question 2
1 / -0

Two identical balls undergo elastic collision Speed of both the balls before colision Is a, maximum possible speed of any ball after collision well be:

A
$$u$$

B
$$2u$$

C
$$\sqrt{3} u$$

D
$$\sqrt{2} u$$

Solution
Question 3
1 / -0

A block of mass M initially at height 'h' on to a spring of force constant k. the maximum compression in the spring is x. then

A
$$ mgh = \frac {1}{2} k x^2 $$

B
$$ mgh(h+x)= \frac {1}{2} k(x+h)^2 $$

C
$$ mgh = \frac {1}{2} k(x+h)^2 $$

D
$$mg(h+x)= \frac {1}{2} k(x)^2 $$

Solution
Question 4
1 / -0

N identical balls are placed on a smooth horizontal surface. Another ball of same mass collides elastically with velocity $$u$$ with first ball of N balls. A process of collision is thus started in which first ball collides with second ball and the second ball with the third ball and so on. The coefficient of resulting for each collision is $$e$$. Find speed of Nth ball :

A
$$(1+e)^{N}u$$

B
$$u(1+e)^{N-1}$$

C
$$\cfrac{u(1+e)^{N-1}}{2^{N-1}}$$

D
$$u^{N}(1+e)^{N-1}$$

Solution
Question 5
1 / -0

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then the energy stored in the wire is

A
0.1 J

B
0.2 J

C
10 J

D
20 J

Solution

Work done in stretching wire

$$=\dfrac {1}{2}\dfrac {YAL^{2}}{L}=\dfrac {1}{2}F.l$$

Where

$$L=$$ length of wire

$$l=$$ increase in length

We know that

$$E=\dfrac {1}{2}\times F \times AL$$

$$=\dfrac {1}{2}\times 200 \times 10^{-3}$$

$$=0.1J$$

Hence $$(A)$$ option is correct.
Question 6
1 / -0

A mass is performing circular motion in a vertical plane centered at $$O$$. The average velocity of the particle is in creased, then at which point the string will break:

A
$$A$$

B
$$B$$

C
$$C$$

D
$$D$$

Solution
Question 7
1 / -0

A force is acting on a particle varies with the displacement x as $$F=ax-bx^2$$. Where a=1 N/m and $$b=1 N/m^2$$. The work done by this force for the first one meter (F is in newtons, x is in meters) is :

A
$$\dfrac{1}{6}J$$

B
$$\dfrac{2}{6}J$$

C
$$\dfrac{3}{6}J$$

D
None of these

Solution
Question 8
1 / -0

The kinetic energy of a particle moving along a circle of radius $$R$$ depends on the distance covered s as $$T={ KS }^{ 2 }$$ where K is a constant. Find the force acting on the particle as a function of $$S$$ -

A
$$\dfrac { 2K }{ S } \sqrt { 1+(\frac { S }{ R } ) } ^{ 2 }$$

B
$$2KS\sqrt { 1+(\frac { R }{ S } ) } ^{ 2 }$$

C
$$2KS\sqrt { 1+(\frac { S }{ R } ) } ^{ 2 }$$

D
$$\dfrac { 2S }{ K } \sqrt { 1+(\frac { S }{ R } ) } ^{ 2 }$$

Solution
Question 9
1 / -0

A body of mass 3 kg is under a constant force, Which causes a displacement s in meter in it, given by the relation $$ s = \frac { 1 }{ 3 } t^2 $$, Where t is in second. Work done by the force in 2 s is.

A
$$ \frac { 5 }{ 19 } J $$

B
$$ \frac { 3 }{ 8 } J $$

C
$$ \frac { 8 }{ 3 } J $$

D
$$ \frac { 19 }{ 5 } J $$

Solution
Question 10
1 / -0

A stone is projected from a horizontal plane. It attains maximum height $$H$$ and strikes a stationary smooth wall and falls on the ground vertically below the maximum height. Assume the collision to be elastic the height of the point on the wall where ball will strike is:

A
$$H/2$$

B
$$H/4$$

C
$$3H/4$$

D
none of these

Solution