Question

A 29.0-kg block is initially at rest on a horizontal surface. A horizontal force of 71.0 N is required to set the block in motion, after which a horizontal force of 56.0 N is required to keep the block moving with constant speed.

Answer 1

(a)
`\mu_s=0.25`

(b)
`\mu_k=0.20`

Step-by-step explanation:

According to Newton's second law:

`\sum F_y:N=mg\\\sum F_x:F_a=F_f`

Recall that the frictional force is related jointly with the coefficient of friction and normal force
`F_f=\mu N`. Replacing in the above equation, we get the coefficient of friction:

`F_a=\mu N=\mu mg\\\mu=(F_a)/(mg)`

(a) The coefficient of static friction is related with the force required to set the block in motion:

`\mu_s=(71N)/(29kg*9.8(m)/(s^2))\\\mu_s=0.25`

(b) The coefficient of kinetic friction is related with the force required to keep the block moving with constant speed:

`\mu_k=(56N)/(29kg*9.8(m)/(s^2))\\\mu_k=0.20`