Work Energy and Power Test

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Work Energy and Power Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

The energy stored in wound watch spring is

A
$$ K.E. $$

B
$$P.E. $$

C
Heat energy

D
Chemical energy

Solution

Energy stored in spring is potential energy, and it is defined as
$$E=\dfrac{1}{2}kx^2$$
where k is spring constant , and x is the extension/compression in spring.
Question 2
1 / -0

Two springs have their force constant as $$ k_1 $$ and $$ k_2( k_1 > k_2) $$ When they are stretched by the same force

A
No work is done in case of both the springs

B
Equal work is done in case of both the springs

C
More work is done in case of second spring

D
More work is done in case of first spring

Solution

Relation between Work done and force is defined as
$$ W = \dfrac {F^2}{ 2k} $$
If both springs are stretched by same force then $$ W \propto \dfrac {1}{k} $$
As $$ k_1 < k_2 $$ therefore $$ W_1 < W_2 $$
i.e. more work is done in case of second spring.
Question 3
1 / -0

A particle moves under the effect of a force $$ F =Cx $$ from $$ x = 0 $$ to $$ x =x_1 $$ The work done process is

A
$$ Cx^2_1 $$

B
$$ \frac {1}{2} Cx^2_1 $$

C
$$ Cx_1 $$

D
$$ Zero $$

Solution

Consider small displacement dx,
work done for this small displacement
$$dW=F.dx$$
$$ W =\int^{x_1}_0 F.dx $$
$$W= \int^{x_1}_0 Cx dx = C [ \dfrac {x^2}{2}]^{x_1}_0$$
$$W = \dfrac {1}{2} Cx^2_1 $$
Question 4
1 / -0

If velocity of a body is twice of previous velocity, then kinetic energy will become

A
$$ 2 $$ times

B
$$ \frac{1}{2} $$ times

C
$$ 4 $$ times

D
$$ 1$$ times

Solution

Kinetic energy=$$ \dfrac{1}{2} mv^2 \therefore K.E. \propto v^2 $$
$$v\rightarrow 2v$$
$$K \rightarrow 2^2=4K$$
If velocity is doubled then kinetic energy will become four times.
Question 5
1 / -0

A spring $$ 40 $$ mm long is stretched by the application of a force. If $$ 10$$ N force required to stretch the spring through $$ 1 $$ mm, then work done in stretching the spring through $$ 40 $$ mm is

A
$$ 84 J $$

B
$$68 J $$

C
$$ 23 $$ J

D
$$ 8 J $$

Solution

we know $$F=kx$$
so $$ k = \dfrac {F}{x} $$
putting values:
$$k= \dfrac {10}{ 1 \times 10^{-3} } = 10^4 N/m $$

$$ W = \dfrac {1}{2} kx^2 = \dfrac {1}{2} \times 10^4 \times ( 40 \times 10^{-3})^2 = 8J $$
Question 6
1 / -0

An object of $$m\ kg$$ with speed of $$v\ m/s$$ strikes a wall at an angle $$\theta$$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

A
$$2mv\cos \theta $$

B
$$2mv\sin \theta $$

C
$$0$$

D
$$2\ mv$$

Solution

$$\vec P_1=mv\sin \theta \hat i-mv\cos \theta \hat j$$
and $$\vec P_2=mv\sin \theta \hat i+mv\cos \theta \hat j$$
So change in momentum
$$\overrightarrow{\Delta P}=\vec P_2-\vec P_1=2mv\cos \theta \hat j, | \Delta \vec P| =2mv\cos \theta$$
Question 7
1 / -0

A body of mass$$ 3 $$kg is under a force, which causes a displacement in it is given by $$ S = \frac {t^3}{3} ( in m ) $$ find the work done by the force in first $$ 2 $$ seconds

A
$$ 2 J $$

B
$$ 3.8 J $$

C
$$ 5.2 J $$

D
$$ 24 J $$

Solution

Given:$$ S = \dfrac {t^3}{3}$$

$$ \therefore dS = t^2 dt $$

$$ a = \dfrac {d^2 S}{ st^2} = \dfrac {d^2}{dt^2} [ \dfrac {t^3}{3} ] = 2t m/s^2 $$

Now work done by the force $$ W = \int^2_0 F.dS = \int^2_0 ma.dS $$

$$ \int^2_0 3 \times 2 t \times t^2 dt = \int^2_0 6t^3 dt = \dfrac {3}{2} [ t^4]^2_0= 24 J $$
Question 8
1 / -0

A spring of spring constant $$ 5 \times 10 N/m $$ is stretched initially by $$ 5 $$ cm form the unstretched position. Then the work required to stretch it further by another $$ 5 Cm $$ is

A
$$ 6. 25 N-m $$

B
$$ 12.50 N-m $$

C
$$ 18.75 N-m $$

D
$$ 25.00 N-m$$

Solution

Work required = change in energy
$$W=\Delta E$$
$$ W = \dfrac {1}{2} k (x^2_2 - x^2_1) = \dfrac {1}{2} \times 5 \times 10^3 ( 10^2 - 5^2 ) \times 10^{-4} $$
$$ = 18.75 J $$
Question 9
1 / -0

The decrease in the potential energy of a ball of mass $$ 20 kg $$ which falls from a height of $$ 50 cm $$ is

A
$$ 968 J $$

B
$$ 98 J $$

C
$$ 1980 J$$

D
None of these

Solution

Decrease in potential energy $$\Delta U =mg(h_2-h_1)$$
$$ \Delta U= mgh =20 \times 9.8 \times 0.5=98 J $$
Question 10
1 / -0

The maximum velocity at the lowest point, so that the string just slack at the highest point in a vertical circle of radius l

A
$$\sqrt{gl}$$

B
$$\sqrt{3gl}$$

C
$$\sqrt{5gl}$$

D
$$\sqrt{7gl}$$

Solution

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 60

Result

Work Energy and Power Test - 60

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SHARING IS CARING

Question 1

$$ K.E. $$

$$P.E. $$

Heat energy

Chemical energy

Solution

Question 2

No work is done in case of both the springs

Equal work is done in case of both the springs

More work is done in case of second spring

More work is done in case of first spring

Solution

Question 3

$$ Cx^2_1 $$

$$ \frac {1}{2} Cx^2_1 $$

$$ Cx_1 $$

$$ Zero $$

Solution

Question 4

$$ 2 $$ times

$$ \frac{1}{2} $$ times

$$ 4 $$ times

$$ 1$$ times

Solution

Question 5

$$ 84 J $$

$$68 J $$

$$ 23 $$ J

$$ 8 J $$

Solution

Question 6

$$2mv\cos \theta $$

$$2mv\sin \theta $$

$$0$$

$$2\ mv$$

Solution

Question 7

$$ 2 J $$

$$ 3.8 J $$

$$ 5.2 J $$

$$ 24 J $$

Solution

Question 8

$$ 6. 25 N-m $$

$$ 12.50 N-m $$

$$ 18.75 N-m $$

$$ 25.00 N-m$$

Solution

Question 9

$$ 968 J $$

$$ 98 J $$

$$ 1980 J$$

None of these

Solution

Question 10

$$\sqrt{gl}$$

$$\sqrt{3gl}$$

$$\sqrt{5gl}$$

$$\sqrt{7gl}$$

Solution

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