Question

Find the coefficient of kinetic friction between a 3.80-kg block and the horizontal surface on which it rests if an 87.0-N/m spring must be stretched by 6.50 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.

Answer 1

μ = 0.15

Step-by-step explanation:

Let's start by using Hooke's law to find the force applied to the block

F = k x

F = 87.0 0.065

F = 5.655 N

Now we use the translational equilibrium relation since the block has no acceleration

∑ F = 0

F -fr = 0

F = fr

the expression for the friction force is

fr = μ N

if we write Newton's second law for the y-axis

N -W = 0

N = W = mg

we substitute

F = μ mg

μ = F / mg

μ =
`(5.655)/(3.8 \ 9.8)`

μ = 0.15