A) 25 m
B) 50 m
C) 75 m
D) 100 m
Correct Answer: B
Solution :
Here,\[u=72\,\,km/hr=72\times \frac{5}{18}=20\,\,m/\sec \] Work done against friction \[=\] Loss in\[K.E.\] \[\Rightarrow \] \[\mu mgs=\frac{1}{2}m{{u}^{2}}\] \[\Rightarrow \] \[s=\frac{{{u}^{2}}}{2\mu g}=\frac{{{(20)}^{2}}}{2\times 0.4\times 10}=\mathbf{50}\,\,\mathbf{m}\].You need to login to perform this action.
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