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This is second of our many quizzes on Kinetic energy. Click on this to access more such competitive quizzes on Mechanics.

Kinetic energy Quiz 2

physics

 $4kg$ of water at $0{}^\circ $ melts to convert $0{}^\circ C$ ice in this process block of mass $1kg$ gets energy from water and starts moving with velocity $''v''$. The possible velocity of block can be,<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ca6dd529-9cde-4df9-bb88-0a83aeecb59e3915408286644637493.png"/> 

The kinetic energy of a block of $1$ kg mass is because of freezing of water. Use energy conservation. 

  Heat lost by water = Kinetic energy gained by block along with $1\text{ kg}$ ice $ \\m\times {{L}_{f\text{usion}}}\text{ }=\dfrac{1}{2}\left( m+M \right)\times {{V}^{2}} \\ \Rightarrow 4\times 3600=\dfrac{1}{2}\left( 4+1 \right)\times {{V}^{2}} \\ \therefore V=24\sqrt{10}\text{ m/s}$

A stone of mass $''m''$ strikes to a rigid wall and loses $50\%$ of its velocity if initially moving with velocity $''V''$. Loss in kinetic energy is?

Kinetic energy of particle is $\dfrac{1}{2}m{{v}^{2}}$, where v is velocity of particle. Calculate kinetic energy at two instance then use formula $\text{Loss in energy}=\dfrac{K.{{E}_{i}}-\dfrac{1}{4}K.{{E}_{i}}}{K.{{E}_{i}}}\times 100$ to calculate percentage change. 

  $\text{Let initial velocity is ''V''} \\ \text{Kinetic energy corresponding equals }\dfrac{1}{2}m{{v}^{2}} \\ \text{Then velocity becomes 50 % of initial i}\text{.e}\text{. }\dfrac{V}{2} \\ \Rightarrow\text{Kinetic energy then becomes}=\dfrac{1}{4}\times \dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{4}K.{{E}_{i}} \\ \Rightarrow\text{Loss in energy}=\dfrac{K.{{E}_{i}}-\dfrac{1}{4}K.{{E}_{i}}}{K.{{E}_{i}}}\times 100 \\ \therefore\text{Loss in energy}=75\%$

A block of man $''m''$ have initial velocity ${{V}_{0}}$ and  it ascends on a rough surface of $\mu =0.2$. Find the maximum distance it can go up. $(\text{Assume length of inclined plane is very large})$ $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ee92fb86-cd3e-4e70-a960-eb9307fd83107291709445402956761.png"/>

Kinetic energy converts in friction energy and potential energy. i.e loss in kinetic energy consumed by friction and gain in gravitational potential energy.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/60b8c8e4-223f-4e75-af0e-cc127c461fce2378015948764206878.png"/>$ \\ $ From free body diagram $ \\ $ Loss in kinetic energy = gain in potential energy $+$ frictional energy $ \\ \Rightarrow \dfrac{1}{2}m{{v}^{2}}=\left( mg\sin \theta +\mu mg\cos\theta  \right)S \\ \therefore S=\dfrac{{{v}^{2}}}{2g\left( \sin \theta +\mu \cos \theta  \right)}$

A stone with mass $''m''$ and velocity $''v''$ just before it strikes the ground. It loses $50\%$ of its velocity after the strike. The maximum height above surface it can reach is $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4b66c67e-28e2-4a67-91da-c024ffa957747934204978595881031.png"/> 

Loss in kinetic energy will be equal to gain in potential energy. Find the kinetic energy at a given instance then calculate potential energy as kinetic energy is converting to potential energy. 

 Whole kinetic energy will convert into potential energy $ \\ \Rightarrow K.E=\dfrac{1}{2}m{{\left( \dfrac{v}{2} \right)}^{2}} \\ \Rightarrow \dfrac{1}{8}m{{v}^{2}} \\ \Rightarrow P.E=mgh \\ \Rightarrow mgh=\dfrac{1}{8}m{{v}^{2}} \\ \therefore h=\dfrac{{{v}^{2}}}{8g}$

 A ball of mass $''m''$ is attached to a string with tension $''T''$ and length. String is fixed to the ceiling. Initially a velocity $''v''$ is given to the ball and it travels in a circular motion and stops at angle $''\theta ''$ as shown in figure. The value of $\theta $ will be<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/472c917d-eb2f-4b2d-b416-67a8880ae0192321915112599828422.png"/>

Use conservation of energy at two points, first a bottom, second at angle $\theta $. Then write kinetic and potential energy at both points. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e769bca2-24d3-4732-99b3-a5baf7cb8a0b8554782010364425533.png"/>$ \\ $ Take $h=0$ at bottom  and when velocity becomes zero, $ h=L\left( 1-\cos \theta  \right) \\ \Rightarrow \dfrac{1}{2}m{{v}^{2}}=mgL\left( 1-\cos \theta  \right) \\ \Rightarrow L=\dfrac{{{v}^{2}}}{2g\left( 1-\cos \theta  \right)} \\ \Rightarrow 1-\cos \theta =\dfrac{{{v}^{2}}}{2gL} \\ \therefore \cos \theta =\left( 1-\dfrac{{{v}^{2}}}{2gL} \right)$