Question

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.

Answer 1

The frictional force
`F_(fri) =` 6.446 N

The acceleration of the block a = 6.04
`(m)/(s^(2) )`

Step-by-step explanation:

Mass of the block = 3.9 kg

`\theta = 40`°

`\mu` = 0.22

(a). The frictional force is given by

`F_(fri) = \mu R_(N)`

`R_(N) = mg \cos \theta`

`R_(N) =` 3.9 × 9.81 ×
`\cos 40`

`R_(N) =` 29.3 N

Therefore the frictional force

`F_(fri) =` 0.22 × 29.3

`F_(fri) =` 6.446 N

(b). Block acceleration is given by

`F_(net) = F - F_(fri)`

F = 30 N

`F_(fri)` = 6.446 N

`F_(net)` = 30 - 6.446

`F_(net)` = 23.554 N

The net force acting on the block is given by

`F_(net) = ma`

23.554 = 3.9 × a

a = 6.04
`(m)/(s^(2) )`

This is the acceleration of the block.