Question

6 A block of mass 4 kg is pulled up a rough slope, inclined at 25° to the horizontal, by means of a rope. The rope lies along the line of the slope. The tension in the rope is 30 N. Given that the acceleration of the block is 2 m s-2 find the coefficient of friction between the block and the plane.​

Answer 1

0.2

Step-by-step explanation:

Draw a free body diagram of the block. There are four forces:

Weight force mg pulling down,

Normal force N pushing normal to the surface,

Friction force Nμ pushing parallel to the surface (downwards),

and tension force T pulling parallel to the surface (upwards).

Sum of forces in the normal direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of forces in the parallel direction:

∑F = ma

T − mg sin θ − Nμ = ma

Nμ = T − mg sin θ − ma

Substitute:

(mg cos θ) μ = T − mg sin θ − ma

μ = (T − mg sin θ − ma) / (mg cos θ)

Plug in values and solve:

μ = (30 N − 4 kg × 9.8 m/s² × sin 25° − 4 kg × 2 m/s²) / (4 kg × 9.8 m/s² × cos 25°)

μ = 0.153

Rounded to one significant figure, the coefficient of friction is 0.2.