Manipal Medical Manipal Medical Solved Paper-2002

  • question_answer
    If the cofficient of friction of a plane inclined at\[45{}^\circ \]is 0.5. Then acceleration of a body sliding freely on it will be:

    A)  \[\frac{9.8}{1\sqrt{2}}m/{{s}^{2}}\]       

    B)  \[\frac{9.8}{\sqrt{2}}m/{{s}^{2}}\]

    C)  \[9.8\text{ }m/{{s}^{2}}\]         

    D)  \[4.8\text{ }m/{{s}^{2}}\]

    Correct Answer: A

    Solution :

     Here: Angle of plane is inclined\[\theta =45{}^\circ \] Coefficient of friction\[\mu =0.5\] The acceleration of the body sliding on the inclined plane is given by \[=g(\sin \theta -\mu \cos \theta )\] \[=9.8(\sin 45{}^\circ -0.5\cos 45{}^\circ )\] \[=9.8\left( \frac{1}{\sqrt{2}}-0.5\times \frac{1}{\sqrt{2}} \right)\] \[=\frac{9.8}{2\sqrt{2}}m/{{s}^{2}}\]


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