An 96 N box initially at rest is pulled by a horizontal rope on a horizontal tab

Question

An 96 N box initially at rest is pulled by a horizontal rope on a horizontal table. The coefficients of kinetic and static friction between the box and the table are 14 and 12, respectively. What is the friction force on this box if the tension in the rope is

Part A

0N?

Express your answer in newtons as an integer.

F = ______ N

Part B

29N?

Express your answer in newtons as an integer.

F = ______ N

Part C

47N

Express your answer in newtons as an integer.

F = ______ N

Part D

49N?

Express your answers in newtons as an integer.

F = ______ N

Part E

150N?

Express your answers in newtons as an integer.

F = ______ N

Explanation / Answer

Friction = The Coefficient of Friction*The Normal Force

ƒ = N

The Normal Force is the reaction of the table on the box which is exactly equal to the weight since the table is horizontal.

Maximum static friction is ƒ = N = (1/2)(96) = 48 N

The friction force will exactly balance the pulling force until static friction is overcome, then the friction force is found using the coefficient of kinetic friction.

ƒ = N = (1/4)(96) = 24 N

The kinetic friction is 24N regardless of how high the pulling force goes. At that point more force will simply make the box accelerate faster.

(A)

Pulling = 0N
Friction = 0N

(B)

Pulling = 29N
Friction = 29N

(C)

Pulling = 47N
Friction = 47N

(D)

Static friction has been overcome, the box is moving, kinetic friction must be used...
Pulling = 49N
Friction = 24N

(E)

Pulling = 150N
Friction = 24N

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