Question

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?

Answer 1

θ = 25.1°

Step-by-step explanation:

Given:-

- The force acting on the block, P = 9 N

- The mass of the block, m = 1.2 kg

- The friction force, Ff = 2.8 N

- The acceleration of block, a = 1 m/s^2

Find:-

Determine the angle between the incline and the horizontal surface?

Solution:-

- We will first make a free body diagram of block up an inclined surface at an angle (θ).

- There are three forces acting on the block:

Weight of block (W) : Acts vertically downward

Applied Force (P) : Acts up-hill

Friction force (Ff) : Acts downhill

- We will develop a coordinate system where ( + x ) is up-hill and ( +y) is normal to the inclined surface:

- First we will resolve/transform the Weight (W) which acts downward into a component that acts down the slope.

W = m*g

Wp = m*g*sin(θ)

Where, g: gravitational constant = 9.81 m/s^2

- Apply second Newton equation of motion for the block:

P - Wp - Ff = m*a

9 - 1.2*9.81*sin(θ) - 2.8 = 1.2*1

1.2*9.81*sin(θ) = 5

sin(θ) = 0.42473

θ = 25.1°

Answer: The slope of the inclined surface is θ = 25.1°

Answer 2

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

Step-by-step 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