Question

A pump lifts 35kg of chocolate from a depth of 110m in 60 sec. How much power does the pump is sucking with?
Givens: Draw a FBD representing the direction of the
influencing force, also note the direction of motion

Answer 1

Assume the chocolate is lifted at constant speed (no acceleration) and ignore friction/losses. Work required = gain in gravitational potential energy, so power = energy / time.

Given: m = 35 kg, h = 110 m, t = 60 s, g ≈ 9.8 m/s².

P = m g h / t = (35)(9.8)(110) / 60
= 37,730 J / 60 s ≈ 628.8 W ≈ 6.29 × 10^2 W.

(Using g = 9.81 m/s² gives ≈ 629.5 W.) In horsepower: ≈ 0.84 hp.

Free-body diagram (ASCII):

↑ F_pump (upward applied force)
┌─────────┐
│ chocolate│ ← displacement (up)
└─────────┘
↓ mg (weight, downward)

Notes: Pump must apply an upward force equal to the weight (F = mg) for constant-speed lifting; the pump does positive work mgh while gravity does negative work. The numeric power above assumes ideal (100% efficient) pumping.