Question

a student attaches a rope to a 20.0 kg box of books. he pulls with force of 90.0 n at a angle of 30.0° with the horizontal. the coefficient of kinetic friction between the box and the sidewalk is 0.500. find the acceleration of the box

Answer 1

acceleration = 0.12 m/s^2

Step-by-step explanation:

Using Second Law of Motion,

F = m*a

• F is net force (N)
• m is mass of the box (kg)
• a is acceleration of the box (m/s^2)

μ = f/Fn

• μ is coefficient of friction
• f is the friction force (N)
• Fn is the nomal force acting on the object (N)

Fn = m*g – 90*sin(30)

Fn = (20)*(9.81) – 90*sin(30)

Fn = 151.2 N

f = μ*Fn

f = (0.5)*(151.2)

f = 75.6 N

F = 90cos(30) – 75.6

F = 2.34 N

a = F/m

a = 2.34/20

a = 0.12 m/s^2