2. Solved Papers
  3. J & K CET Engineering

J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer
    The amount of force that is needed to accelerate a truck of mass \[36000\,kg\]from rest to a velocity of \[60\,km/h\]in \[20\,s\] is

    A)  \[6\,kN\]

    B)   \[30\,kN\]

    C)  \[60\,kN\]

    D)  \[30000\,kN\]

    Correct Answer: B

    Solution :

    Given,   \[v=60\,km/h\] \[=60\times \frac{5}{18}=\frac{50}{3}m/s\] \[v=u+ft\] \[\Rightarrow \] \[\frac{50}{3}=0+f\times 20\] \[\Rightarrow \] \[f=\frac{50}{3\times 20}=\frac{5}{6}m/{{s}^{2}}\] Required force,  \[P=mf\] \[=36000\times \frac{5}{6}\] \[=30000\,N\] \[=30\,\,kN\]


You need to login to perform this action.
You will be redirected in 3 sec spinner