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A box of mass 25 kg starts from rest and slides down an inclined plane 8 m long and 5 m high. It is found to move at the bottom at 7$m{{s}^{-1}}$. What is the force of friction?A) 79.6NB) 96.6NC) 76.6ND) 116.6N

Last updated date: 25th Jun 2024
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Hint: We are given that a box is moving from the top of an inclined plane to the bottom without slipping and any external forces. We can find the force of friction experienced by the box as we are given its final velocity after the motion along the incline.

We know that for any object at a height above the surface has a potential energy by virtue of its height. When the situation is apt, the body converts its potential energy to the kinetic energy which triggers the motion of the body.

In our situation, the box of mass 25kg is moved down an inclined plane which is 8m long and 5m high. We know that the box possesses a potential energy at this height, which can be given as –
\begin{align} & PE=mgh \\ & \Rightarrow PE=25\times 10\times 5 \\ & \Rightarrow PE=1250J \\ \end{align}
We know that for a body undergoing motion on a smooth surface without any fictional forces acting, the total potential energy will be converted to kinetic energy with a velocity. But here there exists a frictional force so is there a work done against the friction which is reduced from the total potential energy. As a result, the potential energy gets converted to both kinetic energy and the work done against friction.
So, we can find the work done against friction as –
\begin{align} & PE=KE+{{W}_{f}} \\ & \Rightarrow {{W}_{f}}=PE-KE \\ & \Rightarrow {{W}_{f}}=mgh-\dfrac{1}{2}m{{v}^{2}} \\ & \text{given,} \\ & v=7m{{s}^{-1}} \\ & \Rightarrow {{W}_{f}}=1250-\dfrac{1}{2}\times 25\times {{7}^{2}} \\ & \Rightarrow {{W}_{f}}=1250-612.5 \\ & \Rightarrow {{W}_{f}}=637.5J \\ \end{align}
Now, we can find the force of friction as the work done per unit distance. The distance travelled by the box is 8m.
i.e.,
\begin{align} & {{F}_{f}}.S={{W}_{f}} \\ & \Rightarrow {{F}_{f}}=\dfrac{{{W}_{f}}}{S} \\ & \Rightarrow {{F}_{f}}=\dfrac{637.5}{8} \\ & \therefore {{F}_{f}}=79.6N \\ \end{align}
So, the force of friction which acted on the box through the inclined plane is 79.6N.

The correct answer is option A.

Note:
The frictional force acting on the box will vary with the normal reaction and the inclination of the plane. In this problem, we are finding the possible frictional force on the box compared with the final velocity of the box at the bottom of the inclined plane.