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

Work and energy Quiz 2

physics

Ram loves to drive cars, one day he went on a long drive with his friends in a car of mass $950 kg$ and the speed of car is changed from $0$ to $5\,m/s$ in $12\,sec$ the total work done by car during this time will be equal to- 

Change in momentum with respect to time of a body is equal to net force exerted on a body. 

  Applying the 3rd equation of motion, we have, $ \\v=u+at\\ \Rightarrow 5=a\times 12\\ \Rightarrow a=\dfrac{5}{12}\\ $ Applying 3rd equation of motion, we have $ \\{{v}^{2}}={{u}^{2}}+2as\\ \Rightarrow {{5}^{2}}=2\times \dfrac{5}{12}s\\ \Rightarrow s=30m\\ $ Now, force is change in momentum and work done is force multiplied by displacement so, $ \\W=\dfrac{m\Delta v}{\Delta t}.s\\ \Rightarrow W=\dfrac{950\times 5}{12}.30\\ \therefore W =11,875\,J$

 Amit went to a park and saw a bird flying above his head at a height of $15\text{ }m$, so he decided to throw a stone of mass $1\text{ kg}$ vertically to find if he can throw a stone up to the height of the bird. The stone reaches a height of $10\text{ }m$ so he wants to calculate the additional kinetic energy given, the next stone so it will reach the required height. The additional kinetic energy will be- 

 The total energy of stone will remain conserved during the motion of the object. 

  Applying energy conservation, we have $ \\K.E=P.E \\ \Rightarrow \dfrac{1}{2}m{{v}^{2}}=mgh\\ \Rightarrow \dfrac{1}{2}m{{v}^{2}}=1\times 10\times 10=100J\\ $ For next cycle it will reach at $15m$ so, $ \\ \Rightarrow K.{{E}_{2}}=mg{{h}_{2}}\\ \Rightarrow K.{{E}_{2}} =1\times 10\times 15\\ \Rightarrow K.{{E}_{2}} =150J\\ $ So, additional kinetic energy will be $ \\ \therefore 150J-100J=50\,J$

A ball of mass $1\text{ }kg$ rolls down from a circular path as shown in the figure to the ground level and eventually comes to rest after some time. Ram sees this phenomenon and discovers that friction on a horizontal portion causes an opposing force and tries to stop the ball. The total work done by friction to stop the ball is calculated by Ram will be-<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0743d49c-82bb-4ba3-8dbe-1edfef5e3a117703053398630543094.png"/> 

 The total potential energy will be converted into kinetic energy after that in some other form of energies due to work done by friction. 

 The total potential energy at the top position will be $ \\P.E=mgh\\ \Rightarrow P.E =1\times 10\times 20\\ \Rightarrow P.E =200J\\ $ This whole energy will converted to different form as it body will comes eventually at rest so the work done by friction will be $200J$

A man of mass $50\text{ }kg$ started running in a marathon from $0\text{ }m{{s}^{-1}}$ to $1\text{ }m{{s}^{-1}}$ in $4\text{ }s$, after some time his velocity becomes $2\text{ }m{{s}^{-1}}$. The work done by man in increasing its velocity from $1\text{ }m{{s}^{-1}}$ to $2\text{ }m{{s}^{-1}}$ will be- 

The amount of work done by the man will be used in increasing the kinetic energy of the man so the difference will be equal to the total work done. 

 Work done will be equal to change in kinetic energy of man so $ \\K.{{E}_{i}}=\dfrac{1}{2}m{{v}^{2}}\\ \Rightarrow K.{{E}_{i}} =\dfrac{1}{2}\times 50\times {{1}^{2}}\\ \Rightarrow K.{{E}_{i}} =25J\\ \Rightarrow K.{{E}_{f}}=\dfrac{1}{2}m{{v}^{2}}\\ \Rightarrow K.{{E}_{f}} =\dfrac{1}{2}\times 50\times {{2}^{2}}\\ \Rightarrow K.{{E}_{f}} =100J\\ $ So work done = change in kinetic energy $ =100-25=75\text{ }\,J$

Ram and Shyam are two friends, one day they were driving cars of mass $500\text{ kg}$ and $450\text{ kg}$ on two parallel roads in same direction with a velocity of $2\text{ }m{{s}^{-1}}$ and $3\text{ }m{{s}^{-1}}$. the total energy will be maximum for which car- 

 The kinetic energy is the only factor for energy on a frictionless road so by comparing them we calculated maximum energy. 

 Only kinetic energy will work on this condition so $ \\K.{{E}_{1}}=\dfrac{1}{2}m{{v}^{2}}\\ \Rightarrow K.{{E}_{1}} =\dfrac{1}{2}\times 500\times {{2}^{2}}\\ \Rightarrow K.{{E}_{1}} =1000J\\ \Rightarrow K.{{E}_{2}}=\dfrac{1}{2}m{{v}^{2}}\\ \Rightarrow K.{{E}_{2}} =\dfrac{1}{2}\times 450\times {{3}^{2}}\\ \Rightarrow K.{{E}_{2}} =2025\,J\\ $ Kinetic energy for Shyam’s car is more than ram.