Work Energy and Power Test

Result

Work Energy and Power Test - 54

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Question 1
1 / -0

A position dependent force $$F = 7 - 2x + 3x^2$$ newton acts on a small body of mass $$2 kg$$ and displaces it from $$x = 0$$ to $$x = 5$$. The work done in joule is :

A
$$70$$

B
$$270$$

C
$$35$$

D
$$135$$

Solution
Question 2
1 / -0

$$\overrightarrow { F } =x\hat i+y\hat j$$, is this force:

A
a conservation force

B
a non conservation

C
depends upon X

D
depends upon $$y \quad \& \quad x$$

Solution

A conservative force depends only on the position of the object. If a force is conservative, it is possible to assign a numerical value for the potential at any point and conversely, when an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken, contributing to the mechanical energy and the overall conservation of energy. If the force is not conservative, then defining a scalar potential is not possible, because taking different paths would lead to conflicting potential differences between the start and end points.

Hence It;s a conservative force.

So, option 'A' is correct
Question 3
1 / -0

A force of $$1200 \mathrm { N }$$ acting on a stone by means of a rope slides the stone through a distance{ of$$ 10 \mathrm { m }$$in a direction inclined at $$ 60 ^ { \circ } $$ force. The work done by the force is

A
$$6000 \sqrt { 3 } J$$

B
6000 J

C
12000 J

D
8000 J

Solution

Hence, the option $$B$$ is the correct answer.
Question 4
1 / -0

$$\begin{array} { l } { \text { A man weighing } 80 \mathrm { kg } \text { climbs a staircase } } \\ { \text { carrying a } 20 \mathrm { kg } \text { load. The staircase has } 40 } \\ { \text { steps, each of } 25 \mathrm { cm } \text { height. If he takes } 20 } \\ { \text { seconds to climb, the work done is } } \end{array}$$

A
$$9800 \mathrm { J }$$

B
$$245 \mathrm { J }$$

C
$$98 \times 10 ^ { 4 } J$$

D
7840 J

Solution

As the man climb up the stair he does the work against the gravity$$,$$ the total gravitational force act on it in downward direction $$=(80+20)g$$
here $$g$$ is the acceleration due to gravity$$,$$
height raised by the man $$25cm=0.25m$$
work done by the man $$\left( {80 + 20} \right) \times g \times 0.25 = 245J$$
Hence,
option $$(B)$$ is correct answer.
Question 5
1 / -0

Five identical balls each of mass m and radius r are strung like beads at random and at rest along a smooth, rigid horizontal thin rod of length L, mounted between immovable supports. Assume $$10r<L$$ and that the collision between balls or between balls and supports are elastic. If one ball is struck horizontally so as to acquire a speed v, the average force felt by the support is

A
$$\dfrac { 5m }{ L-5r } $$

B
$$\dfrac { m{ v }^{ 2 } }{ L-10r } $$

C
$$\dfrac { 5m{ v }^{ 2 } }{ L-10r } $$

D
$$\dfrac { m{ v }^{ 2 } }{ L-5r } $$

Solution
Question 6
1 / -0

The displacement (x) and time (t) for a particle are related as $$t=\sqrt { x } +3$$. What is the work done in the first six second of its motion?

A
3 J

B
4 J

C
Zero

D
5 J

Solution

Given,

$$t=\sqrt x +3$$

Then,

$$\sqrt x=t-3 \implies x=t^2+9-6t$$

Also,

$$v=\dfrac{dx}{dt}= 2t-6$$

At $$t=0, v=-6\,ms^{-1}$$

At $$t=6\,sec, v=6\, ms^{-1}$$

Work done = Change in K.E $$=K.E_f-K.E_i$$

$$=\dfrac 12 m(6)^2-\dfrac 12 m(6)^2=0$$
Question 7
1 / -0

When a 4 kg mass is hung vertically on a light spring obey's Hooks law , the spring stretches by 2 cms. The work required to be done by an external agent in stretching this spring by 5 cm will be ( g =9.8 $${ ms }^{ -2 }$$ )

A
4.9 J

B
2.4 J

C
0.495 J

D
0.24 J

Solution
Question 8
1 / -0

A variable force given by $$ F= ( 3x^2 \hat i + 4 \hat j) $$ acts on a particle. the force is in newton and x is in meter. what is the change in KE of particle as it moves from point (2, 3) to (3, 0)

A
-7J

B
zero

C
7J

D
19J

Solution
Question 9
1 / -0

A particle moves along positive X-axis from x=$$0$$ to x=$$5$$ under influence of force is given by$$F={ x }^{ 2 }-2x-8.$$ The work done by the force is

A
-69 units

B
-23.33 units

C
-5 units

D
zero

Solution
Question 10
1 / -0

Two bodies of masses $$4kg$$ and $$16kg$$ are at rest. The ratio of times for which the same force must act on them to produce the same kinetic energy in both of them is

A
$$1:4$$

B
$$2:1$$

C
$$1:2$$

D
$$4:1$$

Solution

kinetic energy is given by$$:$$
so taking velocity for mass $$1kg$$ as $$v_1,$$ we get
and taking velocity for mass $$4kg$$ as $$v1,$$ we have
given that $$e_1 = e_2, $$
equating the above terms and simplifying, we get$$,$$
now , momentum is given by$$,$$
let us assume $$, p_1$$ as momentum for mass $$1kg$$ and $$p_2$$ for mass $$4kg$$
so $$,$$
$$p_1 = 1 × v_1$$
and $$p_2 = 4×v_2$$
then$$,$$ $$\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{1}{4} = 1:4$$
Hence,
option $$(A)$$ is correct answer.