A 1.8-kg block is released from rest at the top of a rough 30° inclined plane. As the block slides down the in

Question

A 1.8-kg block is released from rest at the top of a rough 30° inclined plane. As the block slides down the incline, its acceleration is 3.0 m/s2 down the incline. Determine the magnitude of the force of friction acting on the block.

4.2 N

3.0 N

3.4 N

3.8 N

2.3 N

Answer 1

Summing forces along the plane,

mg(sin 30) - f = ma

where

m =mass of the block = 1.8 kg (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

f = frictional force

a = acceleration of block = 3 m/sec^2 (given)

Substituting appropriate values,

(1.8)(9.8)(sin 30) - f = (1.8)(3)

Solving for "f",

f = 1.8[(9.8*0.5) - 3]

f = 3.42 N

This is the third option in your given choices.

Answer 2

The force causing the slide is m*g*sinø - Fr, where Fr is the frictional force.  This will cause an acceleration = m*a, so

m*a = m*g*sinø - Fr

Fr =m*g*sinø - m*g