A 1.2 kg block is pushed up a roughb inclined by a force of 9 N acting parallel to the incline. The friction force is 2.8 N. The acceleration of the block is 1 m/s^2 up the incline. Determine the angle between the incline and the horezontal surface?

The angle between the inclined block and the horizontal surface is 25.16⁰

Explanation:

Given;

mass of block, m = 1.2 kg

force acting parallel to the inclined plane, F = 9 N

frictional force on the block, Fk = 2.8 N.

acceleration of the block. a = 1 m/s²

According to Newtons second law of motion, sum of all the forces acting on the block is given by;

∑F = Ma

F - W - N = ma

where;

F is the parallel force on the block, acting upwards

W is the weight of the block inclined at an angle θ, acting downwards

N is the normal reaction on the block, acting upwards = frictional force on the block.

F - mgsinθ - Fk = ma

9 - (1.2 x 9.8)sinθ - 2.8 = 1.2 x 1

6.2 - 11.76sinθ = 1.2

11.76sinθ = 6.2 - 1.2

11.76sinθ = 5

sinθ  = 5 / 11.76

sinθ  = 0.4252

θ = sin⁻¹ (0.4252)

θ = 25.16⁰

Therefore, the angle between the inclined block and the horizontal surface is 25.16⁰

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