Work and Energy Test

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Work and Energy Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

If the mass of the moving object is decreased 1/4 of its mass and its velocity is increased to twice its previous velocity, what will be the kinetic energy of the object from the following?

A
1/2 of the previous kinetic energy

B
4 times of previous kinetic energy

C
Kinetic energy will remain constant

D
2 times of the previous kinetic energy

Solution

We know,
Kinetic energy, $$KE = \dfrac{1}{2}mv^{2}$$
where, $$m=mass$$ and $$v=velocity$$

If the mass of the moving object is decreased $$1/4$$ of its mass and its velocity is increased to twice its previous velocity,
$$m'=\dfrac m4$$

$$v'=2v$$

$$KE' = \dfrac{1}{2}m'v'^{2}$$

$$KE' = \dfrac{1}{2}\left(\dfrac m4\times (2v)^2\right)$$

$$KE' = \dfrac{1}{2}mv^{2}$$

$$KE'=KE$$
Question 2
1 / -0

The energy possessed by a body due to its change in position or shape is called:

A
Potential energy

B
nuclear energy

C
kinetic energy

D
chemical energy

Solution

Energy possessed by a body by virtue of its position or shape is called potential energy
Question 3
1 / -0

A man carries a load on his head through a distance of 5 m. The maximum amount of work is done when he

A
Movies it over an inclined plane

B
Movies it over a horizontal surface

C
Lift it vertically upwards

D
All of the above

Solution

The maximum work done by man will be when he lift it vertically upwards because in such situation the man has to exert force opposite to gravity that is in the direction of the displacement of load.
Question 4
1 / -0

Let us suppose there are two balls of an equal mass shape and size. We apply an equal force on both. Then work done by the force will be:

A
Same

B
Zero

C
Different

D
None of the above

Solution

Given,
Two balls of equal mass, shape, and size.
Mass of each ball, $$m$$
Force applied on each ball, $$F$$
We know,
Work done, $$W=F\times s$$
where, $$s=\text{displacement}$$
As both the balls are identical, the distance traveled by each ball will be the same.
$$\therefore$$ Work done by the force will be the same.
Question 5
1 / -0

An object is thrown vertically upwards. As it rise, its total energy

A
Decreases

B
Increases

C
Remains constant

D
None

Solution

Total Energy = Kinetic Energy + Potential Energy

As object move upward is velocity will decrease due to deceleration by gravity.

So its kinetic energy =$$\dfrac{1}{2}m{{v}^{2}}$$ will decrease as it move upward.

But its potential energy =$$mgH$$ will increase as it move upward.

Hence, total energy will remain constant.
Question 6
1 / -0

A body of mass $$m$$ kg is lifted by a man to a height of one metre in $$30$$ sec.Another man lifts the same mass to the same height in $$60$$ sec. The work done by them are in the ratio

A
$$1 :1$$

B
$$2 : 1$$

C
$$4 : 1$$

D
$$1 : 2$$

Solution

Work done $$W= F\cdot s$$.....(1)
Force applied to pull the object $$F=$$ Wight of the object
$$F= W=mg$$.........(2)
From equation 1 and 2
$$W=mg\cdot s$$ Hight attend by the mass $$=h$$ (let)
$$W=mgh$$ $$s=h$$
Hence Work done is independent of time taken to lift the mass. so
$$W_1:W_2 :: 1:1$$ Option A
Question 7
1 / -0

The $$P.E.$$ and $$K.E.$$ of a helicopter flying horizontally at a height $$400\ m$$ are in the ratio $$5:2$$. The velocity of the helicopter is:

A
$$56\ m/s$$

B
$$28\ m/s$$

C
$$14\ m/s$$

D
$$42\ m/s$$

Solution

Let mass and velocity of the helicopter be $$'m'$$ and $$'v'$$
So,
$$KE=\dfrac{1}{2}mv^2$$ and
$$PE=mgh$$, where $$[h=400\ m]$$
Given ratio,
$$\dfrac{PE}{KE}=\dfrac{5}{2}$$
$$\Rightarrow \dfrac{mgh}{\dfrac{1}{2}mv^2}=\dfrac{5}{2}$$
$$\Rightarrow v^2=\dfrac{4}{5} gh$$.
$$\Rightarrow v=\sqrt{\dfrac{4}{5}\times 10\times 400}$$
$$\Rightarrow v=40\sqrt{2}\simeq 56.7\ m/s$$.
Question 8
1 / -0

A staircase has $$40$$ steps each of width $$35cm$$ and height $$25cm$$.A boy of mass $$20kg$$ ascends the staircase.The work done by him is (taken $$g=10ms^{-2})$$

A
$$2\times 10^5J$$

B
$$4.8\times 10^3J$$

C
$$2\times 10^{3}J$$

D
$$4.8 \times 10^5J$$

Solution

$$ Work done = Force \times displacement$$.
$$d=40\times 25cm = 1000cm$$
$$F=20\times 10 = 200 N$$
$$W=200 N \times 10 m$$
$$=2\times 10^3 J$$
Question 9
1 / -0

A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at $$x=0$$ has the kinetic energy of $$25$$ J, then the kinetic energy of the particle at $$x=16$$m is?

A
$$45$$J

B
$$30$$J

C
$$70$$J

D
$$135$$J
Question 10
1 / -0

An object of mass $$5 \ kg$$ falls from rest through a vertical distance of $$20 \ m$$ and reaches a velocity of $$10 \ m/s$$. How much work is done by the push of the air on the object?

A
$$-740 \ J$$

B
$$-750 \ J$$

C
$$-760 \ J$$

D
$$-770 \ J$$

Solution

From work energy theorem,
work done by weight + work done by air$$=ΔKE$$ (change in Kinetic Energy)
$${ W }_{ air }=ΔKE−{ W }_{ (weight) }$$
$$=\dfrac{1}{2}m{ v }^{ 2 }−mgh$$
$$=\dfrac{1}{2}×5×{ (10) }^{ 2 }−5×10×20$$
$$=−750J$$

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Re-Attempt Weekly Quiz Competition

Work and Energy Test - 46

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Work and Energy Test - 46

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SHARING IS CARING

Question 1

1/2 of the previous kinetic energy

4 times of previous kinetic energy

Kinetic energy will remain constant

2 times of the previous kinetic energy

Solution

Question 2

Potential energy

nuclear energy

kinetic energy

chemical energy

Solution

Question 3

Movies it over an inclined plane

Movies it over a horizontal surface

Lift it vertically upwards

All of the above

Solution

Question 4

Same

Zero

Different

None of the above

Solution

Question 5

Decreases

Increases

Remains constant

None

Solution

Question 6

$$1 :1$$

$$2 : 1$$

$$4 : 1$$

$$1 : 2$$

Solution

Question 7

$$56\ m/s$$

$$28\ m/s$$

$$14\ m/s$$

$$42\ m/s$$

Solution

Question 8

$$2\times 10^5J$$

$$4.8\times 10^3J$$

$$2\times 10^{3}J$$

$$4.8 \times 10^5J$$

Solution

Question 9

$$45$$J

$$30$$J

$$70$$J

$$135$$J

Question 10

$$-740 \ J$$

$$-750 \ J$$

$$-760 \ J$$

$$-770 \ J$$

Solution

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