Worked example 5.5: Sliding down a plane

Question: A block of mass $m=3 {\rm kg}$ starts at rest at a height of $h=43 {\rm cm}$ on a plane that has an angle of inclination of $\theta=35^\circ$ with respect to the horizontal. The block slides down the plane, and, upon reaching the bottom, then slides along a horizontal surface. The coefficient of kinetic friction of the block on both surfaces is $\mu=0.25$. How far does the block slide along the horizontal surface before coming to rest?

Answer: The normal reaction of the plane to the block's weight is

$$R = m g \cos\theta.$$

Hence, the frictional force acting on the block when it is sliding down the plane is

$$f = \mu R = 0.25\times 3\times 9.81 \times \cos 35^\circ = 6.03 {\rm N}.$$

The change in gravitational potential energy of the block as it slides down the plane is

$${\mit\Delta}U = -m g h = - 3\times 9.81 \times 0.43 = -12.65 {\rm J}.$$

The work $W$ done on the block by the frictional force during this process is

$$W = -f x,$$

where $x= h/\sin\theta$ is the distance the block slides. The minus sign indicates that $f$ acts in the opposite direction to the displacement of the block. Hence,

$$W = -\frac{6.03 \times 0.43 }{\sin 35^\circ} =- 4.52 {\rm J}.$$

Now, by energy conservation, the kinetic energy $K$ of the block at the bottom of the plane equals the decrease in the block's potential energy plus the amount of work done on the block:

$$K = - {\mit\Delta}U + W = 12.65 - 4.52 = 8.13 {\rm J}.$$

The frictional force acting on the block when it slides over the horizontal surface is

$$f' = \mu m g = 0.25\times 3\times 9.81 = 7.36 {\rm N}.$$

The work done on the block as it slides a distance $y$ over this surface is

$$W' = -f' y.$$

By energy conservation, the block comes to rest when the action of the frictional force has drained all of the kinetic energy from the block: i.e., when $W'=-K$. It follows that

$$y = \frac{K}{f'} = \frac{8.13}{7.36} = 1.10 {\rm m}.$$