Question

A block of mass m = 5.0 kg is pulled by a constant force across a flat floor so that it moves with a constant acceleration. The force has a magnitude F = 60 N and is directed up at an angle of θ = 30° with respect to horizontal. The block’s bottom surface always remains in contact with the floor. The kinetic friction between the block and floor has a coefficient µk = 0.30.

Calculate the magnitude of the block’s acceleration.

Answer 1

Answer:

The magnitude of the block’s acceleration is 7.45m/s²

Explanation:

The magnitude of the block acceleration can be calculated using the Newton's second law of motion.

Force = mass × acceleration

The body placed on the flat floor is acted upon by number of forces as shown in the attachment.

Taking the sum of forces along the horizontal, we will have;

Fm - Ff = ma... (1) where;

Fm is the moving force (The force causing the object to move)

Fm = 60cos30° = 51.96N

Ff = µkR which is the frictional force (opposing force)

m is the mass = 5.0kg

a is the acceleration of the object

Substituting this values into equation 1 above we will have;

Fm -  µkR = ma

Note that weight W = normal reaction R = mg

Fm - µkmg= ma

51.96 - 0.3(5)(9.81)= 5a

51.96 - 14.72 = 5a

51.96 - 14.72 = 5a

37.25 = 5a

a = 37.25/5

a = 7.45m/s²

The magnitude of the block’s acceleration is 7.45m/s²