Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is the second of our many quizzes on Work-energy principle. Click on this to access more such competitive quizzes on Mechanics.

Work-energy principle Quiz 2

physics

Rahul tried a funny experiment with his friends that can eat the ice as fast as possible in one minute among the group. All ate the ice individually at the same and the winner is Rahul because he ate the ice 50g in a minute so how much amount of work is done to eat the ice by Rahul?$\left( \text{J=4}\text{.2 Jca}{{\text{l}}^{-1}} \right)$ 

If Rahul tries to eat the ice then there must be heat required to melt that ice so we first calculate the heat required then use, $W=JH$. 

  $\text{Mass of the ice (m)= 50g}\\\text{Latent heat of ice, L=80cal/g}\\\text{So, (H) amount of heat will be} = \text{m }\!\!\times\!\!\text{ L=4000 cal}\\\text{Hence, work done will be}\,\text{W=JH=4}\text{.2 }\!\!\times\!\!\text{ 4000=16800J}$

A policeman fires a bullet with a speed of $\text{800m/s}$ on the targeted board and the number of bullets fired by the man is 10 but the policeman did not fire all the bullets correctly on the targeted board. If the weight of one bullet is 50g and work done by the bullets targeted on the board is 64000J then calculates the total number of bullets targeted on the board. 

Use the concept of Work done = Change in kinetic energy. Calculate the number of bullets. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2942b226-60af-43ee-b478-d2a3c5644da15554111627178679527.png"/> <br>$\\ $ Let the number of bullets which actually targeted successfully on the board be $N$ $ \\ $ Mass of the bullets which actually targeted successfully on the board $ =\text{N }\!\!\times\!\!\text{ 0}\text{.050 Kg}\\\text{Velocity=800m/s,Work done=64000J}\\ $ Work done $=$ Change in K.E. $ \\ 64000=\dfrac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}} \\\text{64000=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ N }\!\!\times\!\!\text{ 0}\text{.050 }\!\!\times\!\!\text{ (800}{{\text{)}}^{\text{2}}} \\ $ So, on calculating $N$  $= 4 $ $ \\ $  The number of balls which actually targeted on the board is 4 out of 10.

Harpreet wants to clean the circular pool with an engine. If the height of the circular swimming pool is 40m deep and the diameter of that swimming pool is 6m then how much amount of work is required by the machine to clean the circular swimming pool? Assume that the water present in a pool is up to 35m deep. 

Calculate the average height of the water raised by the motor and also calculate the mass of the water present in the pool then calculate the work done. 

 Radius of the pool $= 3m$ $\\ $ Area of cross section $ = \!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{=}\dfrac{\text{198}}{\text{7}}{{\text{m}}^{\text{2}}}$ $ \\ $ Volume of the water in well $=$ Area of cross section $\times$  depth of the water in pool $\\ = \dfrac{\text{198}}{\text{7}}\text{ }\!\!\times\!\!\text{ 35=990}{{\text{m}}^{\text{3}}} \\ $ Mass of the water $=$ Volume $\times$ density $= 99 \times 10^4Kg$ $ \\ $ As the pool is emptied the height through which water has to be raised by the engine changes from $(40-35)m$ in the beginning to $(45-0)m$ at the end. Hence, Average height raised: $ \\ \dfrac{\text{5+45}}{\text{2}}\text{=25m} \\ $ Workdone $ = mgh = 99 \!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ }\!\!\times\!\!\text{ 10 }\!\!\times\!\!\text{ 25=2475 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{J}$

What is the effect on the velocity of the body, if the time gets 4 times from the initial time and the body is moving with constant power in the same direction?

Use the concept of work energy theorem, $\text{W=P}\times \text{t=}\dfrac{1}{2}m{{v}^{2}}$.

 $\text{According to the work energy theorem,}\\\text{W=P}\times \text{t=}\dfrac{1}{2}m{{v}^{2}}\\{{\text{v}}^{\text{2}}}\text{=}\dfrac{\text{2 }\!\!\times\!\!\text{ P }\!\!\times\!\!\text{ t}}{\text{m}} \\\text{So}\,\text{from}\,\text{above} \\\text{v}\,\text{is directly proportional to }\sqrt{\text{t}} \\\text{So as time becomes 4 times,} \text{The velocity will become 2 times.}$

Alex throws his coin in the upward direction with a velocity of 4m/s and then he again caught that coin, what will be the work done by the gravity on the coin if Alex′s coin mass is only just 150g?

Use the formula: Work done=Loss in K.E. of the body. 

 $\text{Work done=Loss in K.E. of the body}\\\text{Work done=}\dfrac{\text{1}}{\text{2}}\text{m(}{{\text{v}}^{\text{2}}}\text{-}{{\text{u}}^{\text{2}}}\text{)} \\\Rightarrow\text{Work done=}\dfrac{\text{1}}{\text{2}}\text{(0}\text{.150)(}{{\text{0}}^{\text{2}}}\text{-}{{\text{4}}^{\text{2}}}\text{)} \\\therefore\text{Work done =-1}\text{.20J}$