Question

A small block sits at one end of a flat board that is 3.50 m

long. The coefficients of friction between the block and the board are μs
= 0.600 and μk
= 0.400. The end of the board where the block sits is slowly raised until the angle the board makes with the horizontal is α0
, and then the block starts to slide down the board. If the angle is kept equal to α0
as the block slides, what is the speed of the block when it reaches the bottom of the board?
Express your answer with the appropriate units.

Answer 1

The approximate speed of the block when it reaches the bottom of the board is 8.28 m/s.

This can be solved using the equation for the conservation of energy which is:

`\[ PE_{\text{initial}} = KE_{\text{final}} + \text{Work against friction} \]`

From the top of the board that formed the incline, the potential energy of the block is given by the formula:

`\[ PE_{\text{initial}} = mgh \]`

And at the bottom of the board that formed the incline, the kinetic energy of the block is given by:

`\[ KE_{\text{final}} = (1)/(2)mv^2 \]`

Next is the work done against friction, which is given by:

`\[ \text{Work against friction} = \mu_k mgh \]`

Setting up the equation and solving for v, we get:

`\[ mgh = (1)/(2)mv^2 + \mu_k mgh \]`

`\[ gh = (1)/(2)v^2 + \mu_k gh \]`

`\[ v^2 = 2gh - 2\mu_k gh \]`

`\[ v = √(2gh(1 - \mu_k)) \]`

Next is to substitute all the values given in the problem to obtain v:

`\[ v = \sqrt{2 * 9.8 \, \text{m/s}^2 * 3.50 \, \text{m} * (1 - 0.400)} \]`

`\[ v \approx √(2 * 9.8 * 3.50 * 0.600) \]`

`\[ v \approx √(68.6) \]`

`\[ v \approx 8.28 \, \text{m/s} \]`

So, the approximate speed of the block when it reaches the bottom of the board is 8.28 m/s.