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9th Class Science Work and energy Question Bank Work, Energy and Power

  • question_answer
    A car of mass 2000 kg is to be stopped from a speed of 72 km/hr. If the coefficient of friction between tyres and road is 0.4, then distance travelled before coming to rest is \[\left( \text{g }=\text{ 1}0\text{ m}/{{\text{s}}^{\text{2}}} \right)\]

    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\[K.E.\] \[\Rightarrow \]               \[\mu mgs=\frac{1}{2}m{{u}^{2}}\] \[\Rightarrow \]               \[s=\frac{{{u}^{2}}}{2\mu g}=\frac{{{(20)}^{2}}}{2\times 0.4\times 10}=\mathbf{50}\,\,\mathbf{m}\].


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