Question

A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find

(a) the coefficient of static friction.
(b) the coefficient of kinetic friction between the block and the surface.

Answer 1

By using the forces required to initiate and sustain motion along with the weight of the block, the coefficient of static friction is calculated as approximately 0.306 and the coefficient of kinetic friction is approximately 0.245.

Step-by-step explanation:

To solve this physics problem, we need to apply concepts of static and kinetic friction. The coefficient of static friction (\(\mu_s\)) and the coefficient of kinetic friction (\(\mu_k\)) can be found using the force values given that are necessary to initiate and maintain the motion of the block, respectively.

(a) To find the coefficient of static friction, use the formula \(\mu_s = \frac{F_{static}}{N}\), where \(F_{static}\) is the force needed to overcome static friction and \(N\) is the normal force, which is equal to the weight of the block (\(mg\)). Given that \(F_{static} = 75.0\) N and \(N = mg = (25.0\) kg)(9.81 m/s\(^2\)\()), we can calculate the coefficient of static friction.

(b) The coefficient of kinetic friction is calculated similarly using \(\mu_k = \frac{F_{kinetic}}{N}\), where \(F_{kinetic}\) is the force needed to maintain constant speed (60.0 N in this case).

Let's calculate:

1. Normal force, \(N = (25.0\) kg)(9.81 m/s\(^2\)) = 245.25 N.
2. Coefficient of static friction, \(\mu_s = \frac{75.0\) N}{245.25 N} ≈ 0.306.
3. Coefficient of kinetic friction, \(\mu_k = \frac{60.0 N}{245.25 N} ≈ 0.245.

Answer 2

(a) 0.31

(b) 0.245

Step-by-step explanation:

(a)

F' = μ'mg.................... Equation 1

Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.

make μ' the subject of the equation above

μ' = F'/mg............. Equation 2

Given: F' = 75 N, m = 25 kg

constant: g = 9.8 m/s²

Substitute these values into equation 2

μ' = 75/(25×9.8)

μ' = 75/245

μ' = 0.31.

(b) Similarly,

F = μmg.................. Equation 3

Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.

make μ the subject of the equation

μ = F/mg.............. Equation 4

Given: F = 60 N, m = 25 kg, g = 9.8 m/s²

Substitute these values into equation 4

μ = 60/(25×9.8)

μ = 60/245

μ = 0.245