Work Energy and Power Test

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Work Energy and Power Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

A car is moving at $$100\ km/h$$. If the mass of the car is $$950\ kg$$, Its kinetic energy is:

A
$$367\ MJ$$

B
$$0.367\ MJ$$

C
$$3.67\ MJ$$

D
$$3.67\ J$$

Solution

Mass of the car, $$m=950\ kg$$
velocity of the car, $$v= 100\ km/h = { \left( 100\times \dfrac { 5 }{ 18 } \right) }$$m/s.
kinetic energy $$\displaystyle KE=\dfrac { 1 }{ 2 } \times m\times { v }^{ 2 }$$

$$\displaystyle =\frac { 1 }{ 2 } \times 950\times { \left( 100\times \frac { 5 }{ 18 } \right) }^{ 2 }J$$
$$\displaystyle =366512.35J=0.367\quad MJ$$.
Hence, the kinetic energy of the car is 0.367 MJ.
Question 2
1 / -0

A girl weighing 50 kg makes a high jump of 1.2 m.What is her kinetic energy at the highest point? $${(g= 10 ms^{-2}}$$)

A
6000 J

B
600 J

C
60 J

D
Zero

Solution

Given,
mass of girl $$M=50\ kg\\h=1.2\ m$$
A girl is jumping vertically upward, when it will reach at maximum Hight its velocity will become zero
ie $$v_f=0$$
$$K.E=\dfrac12 mv^2=\dfrac12m\times 0=0$$
Option D
Question 3
1 / -0

A rifle bullet loses 1/20th of its velocity in passing through a plank. Assuming constant resistive force, the least number of such planks required just to stop the bullet is :

A
15

B
10

C
11

D
20

Solution

Answer is C.

Let v the initial velocity of the bullet, then
$$\displaystyle \frac { 1 }{ 2 } { mv }^{ 2 }-\frac { 1 }{ 2 } { m\left( \frac { 19v }{ 20 } \right) }^{ 2 }=F.x$$ ...(i)
F= average forces, x = thickness of plank
$$\displaystyle \frac { 1 }{ 2 } { mv }^{ 2 }=Fx.n$$ ...(ii)
Because of work-K.E. theorem, and n = number of blocks
Dividing (ii) by (i), we get n = 10.25
n = 11 (in round numbers)
Hence, the least number of such planks required just to stop the bullet is 11.
Question 4
1 / -0

The moon revolves around the earth because the earth exerts a radial force on the moon. Does the earth perform work on the moon?

A
no

B
yes, sometimes

C
yes, always

D
cannot be decided

Solution

No, the earth does not perform any work on the moon.
Work done(W) is defined as the scalar product of force(F) and displacement(s).
So, W = F $$\ast$$ s = FsCos$$\theta$$ where \theta is the angle between force and displacement vector.
The radial force exerted on the moon by earth i.e attractive force due to gravity acts in direction perpendicular to which the moon suffers the displacement during rotation.
So, $$\theta$$ = 90 hence Cos$$\theta$$ = 0

So, W = |F||s| $$\ast$$ 0 = 0
Hence work done by earth is zero.
Question 5
1 / -0

A ball moving with a velocity v strikes a wall moving toward the ball with a velocity u. An elastic impact lasts for t sec. Then the mean elastic force acting on the ball is

A
$$\displaystyle \frac { 2mv }{ t } $$

B
$$\displaystyle \frac { 2m\left( \upsilon +u \right) }{ t } $$

C
$$\displaystyle \frac { 2m\left( \upsilon +2u \right) }{ t } $$

D
$$\displaystyle \frac { m\left( 2\upsilon +u \right) }{ t } $$

Solution

Relative speed of the ball $$\displaystyle =\left( \upsilon +u \right) $$
Speed after rebouncing $$\displaystyle =-\left( \upsilon +u \right) $$
Now, $$\displaystyle F=m\frac { \Delta \upsilon }{ \Delta t } =\frac { m }{ t } \left[ \left( \upsilon +u \right) \right] -[-\left[ \left( \upsilon +u \right) \right]] $$
$$\displaystyle =\frac { 2m }{ t } \left( \upsilon +u \right) $$
Question 6
1 / -0

Asha lifts a doll from the floor and places it on a table. If the weight of the doll is known, what else does one need to know in order to calculate the work Asha has done on the doll?

A
the time required

B
the height of the table

C
the power she could deliver

D
cost of the doll or the table

Solution

Work done, W = force × displacement = mg × h = mgh
This is the potential energy.
If weight (mg) is known, then we need to know the height(h) of the table.
Question 7
1 / -0

A man throws the brick to the height of 12 m where they reach with a speed of 12 m/s. If he throws the bricks such that they just reach this height, then what percentage of energy will be save?

A
19%

B
38%

C
57%

D
76%

Solution

Case one when final velocity is $$v=12m/s$$

$$W_1=mgh+\dfrac{mv^2}{2}=m\times 10\times 12+\dfrac{m\times 12^2}{2}=192m$$

When Final $$v=0$$
$$W_2=mgh=m\times 12\times 10=120m$$

Percentage Energy saved=$$\dfrac{(192m-120m)\times 100}{192m}=38$$%
Question 8
1 / -0

Two bullets P and Q, masses 10 and 20 g, are moving in the same direction towards a target with velocities of 20 and 10 m/s respectively. Which one of the bullets will pierce a greater distance through the target?

A
P

B
Q

C
Both will cover the same distance

D
Nothing can be decided

Solution

Resistance offered by plank, F = ma
So, F is directly proportional to m
So, less is mass, less resistance, more penetration.
So, P will penetrate more.
Question 9
1 / -0

The relationship between force and position is shown in the figure (in one dimensional case). Work done by the force in displacing a body from
$$ X=1\ cm\ to\ $$$$X=5\ cm\ is$$:

A
700 ergs

B
70 ergs

C
60 ergs

D
20 ergs

Solution

Work is area under the curve.
So $$W=W_1+W_2+W_3+W_4$$
$$W_1=area\ under\ A_1BCM_2 $$
$$W_2=area\ under\ M_2DEF_3 $$
$$W_3=area\ under\ F_3GHI_4 $$
$$W_4=area\ under\ I_4JKL5 $$
$$W_1=10\times 1=10ergs$$
$$W_2=20\times 1=20ergs$$
$$W_3=-20\times 1=-20ergs$$
$$W_4=10\times 1=10ergs$$
$$W=W_1+W_2+W_3+W_4=10+20-20+10=20\ ergs$$
Question 10
1 / -0

A body rolling down a hill has :

A
K.E. only

B
P.E. only

C
neither K.E. nor P.E.

D
both K.E. and P.E

Solution

A ball rolling down a hill has both Kinetic and potential energy. Kinetic energy is due to motion. So if the ball is rolling down it has kinetic energy. Potential energy is the stored energy which is waiting to be used. If the ball is at a greater height it will have more potential energy. So when the body starts rolling the stored potential energy gets converted into kinetic energy which causes its motion. Hence a body rolling down a hill has both K.E and P.E

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 33

Result

Work Energy and Power Test - 33

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SHARING IS CARING

Question 1

$$367\ MJ$$

$$0.367\ MJ$$

$$3.67\ MJ$$

$$3.67\ J$$

Solution

Question 2

6000 J

600 J

60 J

Zero

Solution

Question 3

15

10

11

20

Solution

Question 4

no

yes, sometimes

yes, always

cannot be decided

Solution

Question 5

$$\displaystyle \frac { 2mv }{ t } $$

$$\displaystyle \frac { 2m\left( \upsilon +u \right) }{ t } $$

$$\displaystyle \frac { 2m\left( \upsilon +2u \right) }{ t } $$

$$\displaystyle \frac { m\left( 2\upsilon +u \right) }{ t } $$

Solution

Question 6

the time required

the height of the table

the power she could deliver

cost of the doll or the table

Solution

Question 7

19%

38%

57%

76%

Solution

Question 8

P

Q

Both will cover the same distance

Nothing can be decided

Solution

Question 9

700 ergs

70 ergs

60 ergs

20 ergs

Solution

Question 10

K.E. only

P.E. only

neither K.E. nor P.E.

both K.E. and P.E

Solution

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