Question

A row of five 1.2 N wooden blocks is being pushed across a tabletop at a constant speed by a toy tractor that exerts a force of 1.7 N on the row. What is the coefficient of kinetic friction between the wooden blocks and the tabletop?

Answer 1

To solve this problem we will apply the concepts related to the Kinetic Friction Force for which we define as

`F = \mu_k N`

Where,

N = Normal Force (Mass for gravity)

`\mu_k =` Kinetic frictional coefficient

The total force applied is 1.7N and the Force from the (normal) weight is equivalent to five times 1.2N, therefore:

`F = 5(N)\mu_k`

`1.7 = 5(1.2)(\mu_k)`

`\mu_k = (1.7)/(5*1.2)`

`\mu_k =0.283`

Therefore the coefficient of kinetic friction between the wooden blocks and the tabletop is 0.283