Question

A 8 kg block sits on a ramp inclined at 12 degrees that has a coefficient of friction of 0.5. A string runs from the block over a pulley at the top of the ramp and then hangs straight down to a suspended weight. If the block slides up the ramp at constant velocity, find the mass of the suspended weight.

Answer 1

m = 8 kg

Angle = 12°

Coefficient of friction = 0.5

Force in string = (mass of suspended weight) × g

Normal force = mg cos(Angle)

= 8 × 9.81 × cos(12°)

= 76.765 N

Frictional force = Coefficient of friction × Normal force

= 0.5 × 76.765

= 38.38 N

As the block accelerates upward, Tension in string is greater than frictional force.

Tension in string - Frictional force = ma

As velocity is uniform, a = 0

(mass of suspended weight) × g - 38.38 = 0

Mass of suspended weight = 38.38/9.81

= 3.912 kg