Physics Math

Question

Mary applies a force of 76 N to push a box with an acceleration of 0.60 m/s^2.

When she increases the pushing force to 85 N, the box's acceleration changes to 0.87 m/s^2.

There is a constant friction force present between the floor and the box.

A.) What is the mass of the box in kilograms?

 

B.) What is the coefficient of kinetic friction between the floor and the box?

 

Answer 1

(A)

 

F = applied force

f = friction force

F - f = net force = ma

F₁ = 76 N

a₁ = 0.60 m/s²

F₂ = 85 N

a₂ = 0.87 m/s²

 

F₁ - f = ma₁

F₂ - f = ma₂

 

Subtract bottom equation from top equation.

 

F₁ - F₂ = m(a₁ - a₂)

 

m = (F₁ - F₂)/(a₁ - a₂) = (76 - 85)/(0.60 - 0.87) kg ≅ 33.3 kg

 

(B)

 

f = μmg

 

F₁ - μmg = ma₁ ⇒

 

μ = (F₁ - ma₁)/(mg) = [76 - 33.3(0.60)]/[33.3(9.8)] ≅ 0.172

 

You can also get μ from

 

μ = (F₂ - ma₂)/(mg) = [85 - 33.3(0.87)]/[33.3(9.8)] ≅ 0.172