Work Energy and Power Test

Result

Work Energy and Power Test - 40

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Question 1
1 / -0

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

A
570 mJ

B
450 mJ

C
490 mJ

D
528 mJ

Solution

$$\dfrac{dx}{dt}=3-8t+3t^2$$
$$dx=(3-8t+3t^2)dt$$
$$\dfrac{d^2}{dx^2}=a=-8+6t$$
Force $$F=ma=m(-8+6t)$$
Work done $$F.dx$$
$$F.(3-8t+3t^2)dt$$
$$\int _{ o }^{ w }{ dw\quad =\int _{ 0 }^{ 4 }{ m\left( -8+6t \right) \left( 3-8t+3{ t }^{ 2 } \right) } } dt\\ w=m\int _{ 0 }^{ 4 }{ \left( -24+82t-72{ t }^{ 2 }+18{ t }^{ 3 }\quad \right) dt } \\ =m\overset { 4 }{ \underset { 0 }{ \left[ -24t+41{ t }^{ 2 }-24{ t }^{ 3 }+\dfrac { 3 }{ 2 } { t }^{ 4 } \right] } } \\ =m(176)\\ =3\times { 10 }^{ -3 }\times 176\\ =528\quad mJ$$
Question 2
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When a rubber -band is stretched by a distance x,it exerts a restoring force of magnitude $$F= ax + bx^2$$ where a and b are constants. the work done is stretching the unstretched rubber band by L is :

A
$$\dfrac {aL^{2}}{2}+\dfrac{bL^{3}}{3}$$

B
$$\dfrac{1}{2} \begin {pmatrix} \dfrac{aL^2}{2} + \dfrac{bL^3}{3}\end {pmatrix} $$

C
$$aL^2 + bL^3$$

D
$$\dfrac {1}{2} ( aL^2 + bL^3)$$

Solution
Question 3
1 / -0

A triangular block ABC of mass m and sides 2a lies on a smooth horizontal plane as shown in the figure. Three point masses of mass m each strike the block at A, B and C with speeds v as shown. After the collision, the particles come to rest. Then the angular velocity acquired by the triangular block is (I is the moment of inertia of the triangular block about G, perpendicular to the plane of the block)

A
$$\displaystyle 2mva \frac{(1 + \sqrt{3})}{\sqrt{3} l}$$ clockwise

B
$$\displaystyle \frac{2mva}{ l}$$ clockwise

C
$$\displaystyle \frac{2 \sqrt{3} mva}{ l}$$ clockwise

D
None of these.

Solution

$$h=\dfrac { 2 }{ 3 } \sqrt { 3 } a\dfrac { 2 }{ \sqrt { 3 } } a\\ I\omega =3mvh\\ =2\sqrt { 3 } mva\\ \omega =\dfrac { 2\sqrt { 3 } mva }{ l } $$
Question 4
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A simple pendulum of length $$L$$ carries a bob of mass $$m$$. When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal the net force on the bob is:

A
$$\sqrt { 10 } mg$$

B
$$\sqrt { 5 } mg$$

C
$$4mg$$

D
$$1mg$$

Solution

V=minimum horizontal speed= $$\sqrt { 5Lg } $$
using energy conservation,
$$\frac { 1 }{ 2 } m{ v }^{ 2 }-\frac { 1 }{ 2 } m{ u }^{ 2 }=mgL\\ \therefore \quad \frac { 1 }{ 2 } m{ u }^{ 2 }=\frac { 1 }{ 2 } m{ v }^{ 2 }-mgL\\ =\frac { 1 }{ 2 } m{ 5gL }-mgL\\ \therefore \quad { v }^{ 2 }=\frac { 3gL }{ 2 } \times 2=3gL\\ T=\frac { m{ v }^{ 2 } }{ L } =3mg\\ \therefore \quad force=\quad \sqrt { { \left( 3mg \right) }^{ 2 }+{ \left( mg \right) }^{ 2 } } \quad \\ =\sqrt { 10 } mg$$
Question 5
1 / -0

When a rubber-band is stretched by a distance $$x$$, it exerts a restoring force of magnitude $$F=ax+b{x}^{2}$$ where $$a$$ and $$b$$ are constants.The work done in stretching the unstretched rubber band by $$L$$ is:

A
$$\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { { bL }^{ 3 } }{ 3} $$

B
$$\cfrac { 1 }{ 2 } \left( \cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { { bL }^{ 3 } }{ 2 } \right) $$

C
$$a{ L }^{ 2 }+{ bL }^{ 3 }$$

D
$$\cfrac { 1 }{ 2 } \left( a{ L }^{ 2 }+{ bL }^{ 3 } \right) $$

Solution

Work done by streching dx length is,
$$dw=Fdx$$
$$=(ax+bx^2)dx$$
$$w=\int _{ 0 }^{ L }{ { \left( ax+b{ x }^{ 2 } \right) }dx } \\ =\overset { L }{ \underset { 0 }{ \left[ \dfrac { { ax }^{ 2 } }{ 2 } +\dfrac { { bx }^{ 3 } }{ 3 } \right] } } \\ =\dfrac { { aL }^{ 2 } }{ 2 } +\dfrac { { bL }^{ 3 } }{ 3 } $$
Question 6
1 / -0

A body of mass $$0.5\ kg$$ travels in a straight line with velocity $$v = ax^{1/2}$$ where $$a = 4 m^{}s^{-2}$$. The work done by the net force during its displacement from $$x = 0$$ to $$x = 2\ m$$ is

A
$$1.5\ J$$

B
$$50\ J$$

C
$$10\ J$$

D
None of these

Solution
Question 7
1 / -0

The diagram below shows the path taken by a ball when Sundram kicks it. The potential energy of the ball is highest at ______________

A
P

B
Q

C
R

D
S

Solution
Question 8
1 / -0

When the speed of a body is doubled, its kinetic energy becomes

A
Double

B
Half

C
Quadruple

D
One-fourth

Solution

The kinetic energy of a body is given by:
$$KE=\dfrac{1}{2}mv^2$$

Hence, Kinetic energy depends on the velocity as:
$$KE\propto v^2$$

So, if we double the velocity, then $$KE$$ becomes four times.

Hence, option C is correct.
Question 9
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A particle is tied to one end of a light inextensible string and is moved in a vertical circle, the other end of the string is fixed at the centre. Then for a complete motion in a circle, which is correct.
(air resistance is negligible).

A
Acceleration of the particle is directed towards the centre

B
Total mechanical energy of the particle and earth remains constant

C
Tension in the string remains constant

D
Acceleration of the particle remains constant

Solution

At any time force acting on particle vary and hence acceleration (net) will have different direction at different times. Tension also changes and its minimum at top point. Magnitude of acceleration also varies.
Considering earth and particle as a system and no external force on system is acting, total mechanical energy will be conserved.
Question 10
1 / -0

Which of the following statements is incorrect?

A
Kinetic energy may be zero, positive or negative

B
Power, energy and work are all scalars

C
Potential energy may be zero, positive or negative

D
Ballistic pendulum is a device for measuring the speed of bullets

Solution

The kinetic energy of a body of mass m which is moving with velocity v is $$ K = \dfrac{1}{2}mv^2$$
Mass is always positive and if the velocity is either positive or negative, the kinetic energy is always positive due to the square of velocity.