Work Energy and Power Test

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Work Energy and Power Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $$ t $$ is proportional to

A
$$ t^{1/2} $$

B
$$ t^{3/4} $$

C
$$ t^{3/2} $$

D
$$ t^2 $$

Solution

$$ P = Fv = mav = m( \dfrac {dv}{dt} ) v \Rightarrow \dfrac {P}{m}dt = vdv $$

$$ \Rightarrow \dfrac {P}{m} \times t = \dfrac {v^2}{2} \Rightarrow v = ( \dfrac {2P}{m})^{ 1/2} (t)^{1/2} $$now

$$ s = \int v dt = \int ( \dfrac {2P}{m})^{1 /2} t^{1/2}dt $$

$$ \therefore s = ( \dfrac {2P}{m})^{ 1/2} [ \dfrac { 2t^{ 3/2} }{3} ] \Rightarrow s \propto t^{ 3/2} $$
Question 2
1 / -0

The relation between the displacement $$ X $$ of an object produced by the application of the variable force $$ F $$ is represented by a graph shown in the figure. If the object undergoes a displacement from $$ X=0.5m $$ to $$ X=2.5m $$ the work done will be approximately equal to

A
$$ 16 J $$

B
$$ 32 J $$

C
$$ 1.6 J $$

D
$$ 8 J $$

Solution

Work done = Area under curve and displacement axis
= Area of trapezium
$$ = \dfrac {1}{2} \times $$ ( sum f two parallel lines ) $$ \times $$ distance between them
$$ = \dfrac {1}{2} ( 10 +4) \times ( 2.5 -0.5) $$
$$ = \dfrac {1}{2} 14 \times 2 = 14 J $$
As the area actually is not trapezium so work done will be more than $$ 14 $$ J i.e. approximately $$ 16 $$ J
Question 3
1 / -0

The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation $$ t = \sqrt { x} +3 $$ where x is in meters and t is in seconds . the work done by the force in the first $$ 6 $$ second is

A
$$ 9 J $$

B
$$ 6 J $$

C
$$ 0 J $$

D
$$ 3 J $$

Solution

According to the question,
$$ x =( t-3)^2 \Rightarrow v = \dfrac {dx}{dt} = 2(t -3) $$

at $$ t = 0 ; v_1 = -6 m/s $$ and at $$ t =6 sec , v_2 = 6 m/s $$

So change in kinetic energy $$ = W = \dfrac {1}{2} mv^2_2 - \dfrac {1}{2} mv^2_1 = 0 J$$
Question 4
1 / -0

The force constant of a wire is k and that of another wire is $$2k $$When both the wires are stretched through same distance, then the work done.

A
$$W_2 = 2W^2_1 $$

B
$$ W_2 =2W_1 $$

C
$$ W_2 =W_1 $$

D
$$ W_2 = 0.5 W_1 $$

Solution

Work required = change in energy
$$W=\Delta E$$
$$ W =\frac {1}{2} kx^2 $$
If both wires are stretched through same distance then $$ W \propto k $$ as $$ k_2 = 2k_1 $$ so $$ W_2 = 2W_1 $$
Question 5
1 / -0

A body of mass $$ 0.1 $$ kg moving with a velocity of $$ 10$$ m/s hits a spring (fixed at the other end) of force constant $$ 1000 $$N/m and comes to rest after compressing the spring. The compression of the spring is

A
$$ 0.01 m $$

B
$$ 0.1 M $$

C
$$ 0.2 m $$

D
$$ 0.5 m $$

Solution

From Energy conservation we can say that
The kinetic energy of mass is converted into potential energy of a spring
$$KE=PE$$
$$ \dfrac {1}{2} mv^2 = \dfrac {1}{2} kx^2 $$
$$ \dfrac {1}{2} mv^2 =\dfrac {1}{2} kx^2 \Rightarrow x = v \sqrt { \dfrac {m}{k} } = 10 \sqrt { \dfrac {0.1}{1000}}= 0.1 m $$
Question 6
1 / -0

The spring will have maximum potential energy when

A
it is pulled out

B
it is compressed

C
both (a) and (b)

D
neither (a) nor (b)

Solution

The spring will have maximum potential energy when it is pulled out or compressed in.
Question 7
1 / -0

When spring is compressed its potential energy........

A
Remains constant

B
Reduces

C
Increases

D
Nothing can be said about it.

Solution

When a spring is compressed or stretched, potential energy energy of the spring increases in both the cases. This is because work is done by us in compression as well as stretching.
Question 8
1 / -0

Masses of two bodies are $$1$$ kg and $$4$$ kg respectively. If their kinetic energies are in $$2:1$$ proportion, the ratio of their speeds is .........

A
$$2\sqrt{2}:1$$

B
$$1:\sqrt{2}$$

C
$$1:2$$

D
$$2:1$$

Solution

We know that,
$$K.E.=\dfrac{1}{2}mv^2$$
$$m_1=1kg$$ $$m_2=4kg$$

$$\therefore \dfrac{K_1}{K_2}=\Bigg(\dfrac{m_1}{m_2}\Bigg)\Bigg(\dfrac{v_1}{v_2}\Bigg)^2$$

$$\Rightarrow \dfrac{2}{1}=\Bigg(\dfrac{1}{4}\Bigg)\Bigg(\dfrac{v_1}{v_2}\Bigg)^2$$

$$\Rightarrow \Bigg(\dfrac{v_1}{v_2}\Bigg)^2=\dfrac{8}{1}$$

$$\Rightarrow \Bigg(\dfrac{v_1}{v_2}\Bigg)=\dfrac{2\sqrt{2}}{1}$$
Question 9
1 / -0

A pile driver drives postes into the ground by repeatedly dropping a heavy object on them. Assume the object is dropped from the same height each time. By what factor does the energy of the pile driver-Earth system change when the mass of the object being dropped is doubled?

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

B
$$1$$; the energy is the sane

C
$$2$$

D
$$4$$

Solution
Question 10
1 / -0

A $$10.0-g$$ bullet is fired into a $$200-g$$ block of wood at rest on a horizontal surface. After impact, the block slides $$8.00\ m$$ before coming to rest. If the coefficient of friction between the block and the bullet before impact?

A
$$106\ m/s$$

B
$$166\ m/s$$

C
$$226\ m/s$$

D
$$286\ m/s$$

E
none of those answer is correct

Solution

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Work Energy and Power Test - 61

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Work Energy and Power Test - 61

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SHARING IS CARING

Question 1

$$ t^{1/2} $$

$$ t^{3/4} $$

$$ t^{3/2} $$

$$ t^2 $$

Solution

Question 2

$$ 16 J $$

$$ 32 J $$

$$ 1.6 J $$

$$ 8 J $$

Solution

Question 3

$$ 9 J $$

$$ 6 J $$

$$ 0 J $$

$$ 3 J $$

Solution

Question 4

$$W_2 = 2W^2_1 $$

$$ W_2 =2W_1 $$

$$ W_2 =W_1 $$

$$ W_2 = 0.5 W_1 $$

Solution

Question 5

$$ 0.01 m $$

$$ 0.1 M $$

$$ 0.2 m $$

$$ 0.5 m $$

Solution

Question 6

it is pulled out

it is compressed

both (a) and (b)

neither (a) nor (b)

Solution

Question 7

Remains constant

Reduces

Increases

Nothing can be said about it.

Solution

Question 8

$$2\sqrt{2}:1$$

$$1:\sqrt{2}$$

$$1:2$$

$$2:1$$

Solution

Question 9

$$\dfrac{1}{2}$$

$$1$$; the energy is the sane

$$2$$

$$4$$

Solution

Question 10

$$106\ m/s$$

$$166\ m/s$$

$$226\ m/s$$

$$286\ m/s$$

none of those answer is correct

Solution

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