Work Energy and Power Test

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Work Energy and Power Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

A force $$F=(3{x}^{2}+2x-7)N$$ acts on a $$2kg$$ body as a result of which the body gets displaced from $$x=0$$ to $$x=5m$$. The work done by the force will be:

A
$$5J$$

B
$$70J$$

C
$$115J$$

D
$$270J$$

Solution

Given, $$=3x^2+2x-7, mass=2kg$$

So the work-done $$W=\int_{0}^{5}(3x^2=2x-7) dx=[x^3+x^2-7x]_{0}^{5}=[5^3+5^2-7\times5=115J$$
Question 2
1 / -0

A position dependent force F = $$3x^{2} - 2x + 7$$ acts on a body of mass 7 kg and displace it from $$x$$ = 10 m to $$x$$ = 5 m. The work done on the body is $$x'$$ joule. If both $$F$$ and $$x'$$ are measured in SI units, the value of $$x'$$ is :

A
-835

B
235

C
335

D
935

Solution

Given that
$$F= 3x^2-2x+7$$

Mass of body = $$7\ kg$$

Work done by force, $$W=\int_{10}^5 F\cdot dx=\int_{10}^5(3x^2-2x+7)dx= [x^3- x^2+7x]_{10}^5= -835\ J $$
Question 3
1 / -0

Two springs $$A$$ and $$B(k_{A}=2 k_{B})$$ are stretched by a applying forces of equal magnitudes at the four ends. If the energy stored in $$A$$ is $$E$$, that in $$B$$ is

A
$$E/2$$

B
$$2E$$

C
$$E$$

D
$$E/4$$.

Solution
Question 4
1 / -0

A particle moves along the $$x-$$axis from $$x=0$$ to $$x=5m$$ under the influence of a force given by $$F=$$$$7 - 2x + 3{x^2}N$$. The work done in the process is

A
$$107\ J$$

B
$$270\ J$$

C
$$100\ J$$

D
$$135\ J$$

Solution

$$W=\int_{0}^{5}Fdx$$

$$=\int_{0}^{5}(7-2x+3x^2)dx$$

$$=35-25+125$$

$$=135J$$
Question 5
1 / -0

A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of $$0.1 \ weber / m^2$$. the moment of work done in rotating it through $$180^\circ$$ from its equilibrium position will be

A
0.1 J

B
0.2 J

C
0.4 J

D
0.8 J

Solution

Magnetic moment of current carrying loop $$ = M = iAN = i\left( {\pi {r^2}} \right)N$$
$$W=MB \left( {\cos {\theta _1} - \cos {\theta _2}} \right) = MB\left( {\cos {0^0} - \cos {{180}^0}} \right) = 2MB$$
$$=0.1 J$$
$$\therefore$$ Option $$A$$ is correct.
Question 6
1 / -0

A string of length $$L$$ and force constant $$K$$ is stretched to obtain extension $$l$$. It is further stretched to obtain extension $${l}_{1}$$.The work done in second stretching is

A
$$\cfrac{1}{2}k{l}_{1}(2l+{l}_{1})$$

B
$$\cfrac{1}{2}K{l}_{1}^{2}$$

C
$$\cfrac { 1 }{ 2 } K\left( { l }^{ 2 }-{ l }_{ 1 }^{ 2 } \right) $$

D
$$\cfrac { 1 }{ 2 } K\left( { 2{ l }_{ 1 }^{ 2 }-l }^{ 2 } \right) $$

Solution

Work done= Final total energy-Initial Total energy
Let us assume difference in first and second extention is $$l-l_1$$
Workdone$$=\dfrac{Kl^2}{2}-\dfrac{Kl_1^2}{2}$$
$$=\dfrac{1}{2}K(l^2-l_1^2)$$
Question 7
1 / -0

A boy is swinging on a swing such that his lowest and highest positions are at heights of $$2m$$ and $$4.5m$$ respectively. His velocity at the lowest position is:

A
$$2.5{ms}^{-1}$$

B
$$7{ms}^{-1}$$

C
$$14{ms}^{-1}$$

D
$$20{ms}^{-1}$$

Solution

The height of the boy falls with the swing,
$$h=4.5-2=2.5m$$
Therefore,
The velocity at the lowest position $$=\sqrt{2gh}$$
$$v=\sqrt{2\times9.8\times2.5}=\sqrt{49}$$
$$v=7 m/s$$
Question 8
1 / -0

In the figure, a 4.0 kg ball is on the end of a 1.6 m rope that is fixed at O. The ball is held at point A, with the rope horizontal is given an initial downward velocity. The ball moves theough three qyarters of a circle with no friction and arrives at B, with the rope barely under tension. Thee initial velocity of the ball, at point A, is closest to

A
4.0 m/s.

B
5.6 m/s.

C
6.2 m/s

D
6.8 m/s

Solution

The motion of the ball is always perpendicular to the tension in the Rope. Therefore, the tension does no work on the system and the total mechanical energy of the ball is conserved.
$$E_A = E_B$$
$$mgR + \dfrac{1}{2} mv_A^2 = mg(2R) + \dfrac{1}{2} mv_B^2$$
$$v_A^2 = 2gR +v_B^2$$
Newton's second law for the ball at point B
$$mg + T = \dfrac{mv_B^2}{R}$$
$$T \approx 0$$
$$v_B = gR$$
$$v_A = 3gR$$
= 3(9.8)(1.6)
= 6.9 m/s
Since the velocity is near to 6.8m/s.
Hence (D) option is correct answer
Question 9
1 / -0

A body mass of $$ 6kg $$ is under a force which causes displacement in it given by$$ = \dfrac{{{t^2}}}{4}$$ metres where $$t$$ is time.The work done by the force in $$2$$ seconds is

A
$$12J$$

B
$$9J$$

C
$$6J$$

D
$$3J$$

Solution
Question 10
1 / -0

The velocity $$(v)$$ of a particle of mass $$m$$ moving along x-axis is given by $$v = \alpha \sqrt {x}$$, where $$\alpha$$ is a constant. Find work done by force acting on particle during its motion from $$x = 0$$ to $$x = 2m$$.

A
$$m\alpha^{2}$$

B
$$m\alpha$$

C
$$\dfrac {m\alpha}{2}$$

D
None of these

Solution

$$v=\alpha\sqrt{x}$$
$$\cfrac{dx}{dt}=\alpha\sqrt{x}$$
$$\Rightarrow \int_{0}^{x}{\cfrac{dx}{\sqrt{x}}}=\int_{0}^{t}{\alpha dt}$$
$$x=\cfrac{\alpha^2t^2}{4}$$
$$v=\cfrac{\alpha^2t}{2}$$
$$a=\cfrac{\alpha^2}{2}$$
$$F=ma=\cfrac{m\alpha^2}{2}$$
$$W=F\cdot \Delta x=\cfrac{m\alpha^2}{2}\times 2=m\alpha^2$$

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 45

Result

Work Energy and Power Test - 45

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SHARING IS CARING

Question 1

$$5J$$

$$70J$$

$$115J$$

$$270J$$

Solution

Question 2

-835

235

335

935

Solution

Question 3

$$E/2$$

$$2E$$

$$E$$

$$E/4$$.

Solution

Question 4

$$107\ J$$

$$270\ J$$

$$100\ J$$

$$135\ J$$

Solution

Question 5

0.1 J

0.2 J

0.4 J

0.8 J

Solution

Question 6

$$\cfrac{1}{2}k{l}_{1}(2l+{l}_{1})$$

$$\cfrac{1}{2}K{l}_{1}^{2}$$

$$\cfrac { 1 }{ 2 } K\left( { l }^{ 2 }-{ l }_{ 1 }^{ 2 } \right) $$

$$\cfrac { 1 }{ 2 } K\left( { 2{ l }_{ 1 }^{ 2 }-l }^{ 2 } \right) $$

Solution

Question 7

$$2.5{ms}^{-1}$$

$$7{ms}^{-1}$$

$$14{ms}^{-1}$$

$$20{ms}^{-1}$$

Solution

Question 8

4.0 m/s.

5.6 m/s.

6.2 m/s

6.8 m/s

Solution

Question 9

$$12J$$

$$9J$$

$$6J$$

$$3J$$

Solution

Question 10

$$m\alpha^{2}$$

$$m\alpha$$

$$\dfrac {m\alpha}{2}$$

None of these

Solution

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