A 5.55 kg block covered in sandpaper is pushed along the ceiling of a room under construction. The block is pushed across the ceiling with a force of 77.9 N directed at an angle of 71.5° to the horizontal. If the coefficient of kinetic friction between the paper and the ceiling is 0.574, what acceleration does the block undergo?

Answer 1

The key to solving this kind of problems is to find the net force acting on the block, free of weight (mg) and frictional forces.

Consider forces acting on the block along the horizontal and vertical directions.

Let
force applied on the block, F = 77.9 N
angle with horizontal, θ = 71.5°
Coeff. of kinetic friction, μk=0.574
mass of block, m = 5.55 kg
acceleration due to gravity, g = 9.80
Normal reaction on the block from the ceiling, N

Let's sum forces in the vertical direction first:
upward component of force - weight = 0
Fsinθ - mg - N= 0
from which N can be found.

Consider horizontal forces, positive in direction of F.

net horizontal force, H = Fcosθ - μN

Using Force = mass * acceleration, we conclude that the horizontal acceleration of the block is given by
H = ma
where a= acceleration in the direction of H.

Can you substitute the numbers and work out the answer?

Answer 2

77.9Nsin71.5 - (5.55kg)(9.8m/s^2) - N = 0
N = 19.5N

H = 77.9cos71.5 - (0.574)(19.5N)= 13.5N
a = 13.53 N/5.55 kg = 2.44 m/s^2

Answer 3

Correct!