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ICSE Class 10 Physics 2013 Paper

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Question : 6 of 70
Marks: +1, -0
A force is applied on a body of mass 20 kg20\text{ kg} moving with a velocity of 40 ms−140\text{ ms}^{-1}. The body attains a velocity of 50 ms−150\text{ ms}^{-1} in 2 seconds. Calculate the work done by the body.
Solution:  
   Force     =m  v−ut=   m.a.   \;\text{ Force }\;\;=m\;\frac{v-u}{t}=\;\text{ m.a. }\;
   Displacement     =  v2−u22a\;\text{ Displacement }\;\;=\;\frac{v^2-u^2}{2a}
   Work done     =   Force   ×   Displacement   \;\text{ Work done }\;\;=\;\text{ Force }\;\times\;\text{ Displacement }\;
  =   m.a.     v2−u22a\;=\;\text{ m.a. }\;\;\frac{v^2-u^2}{2a}
W  =  12m(v2−u2)W\;=\;\frac{1}{2}m(v^2-u^2)
W  =  12×20×(502−402)W\;=\;\frac{1}{2}\times20\times(50^2-40^2)
  =  12×20×900\;=\;\frac{1}{2}\times20\times900
  =9000 Joule.\;=9000\text{ Joule}.
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