A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of kinetic friction between the block and the floor is 0.22. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.

The frictional force   6.446 N

The acceleration of the block a = 6.04

Explanation:

Mass of the block = 3.9 kg

°

= 0.22

(a). The frictional force is given by

3.9 × 9.81 ×

29.3 N

Therefore the frictional force

0.22 × 29.3

6.446 N

(b). Block acceleration is given by

F = 30 N

= 6.446 N

= 30 - 6.446

= 23.554 N

The net force acting on the block is given by

23.554 = 3.9 × a

a = 6.04

This is the acceleration of the block.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s