Using a stationary block, applying a force of 0.05 kW, the load is lifted at a speed of 0.1 m / s for 30 minutes.

Using a stationary block, applying a force of 0.05 kW, the load is lifted at a speed of 0.1 m / s for 30 minutes. What kind of work is being done?

Given: F (applied force) = 0.05 kN = 50 N; V (constant lifting speed) = 0.1 m / s; t (duration of lifting the load) = 30 min (in the SI system t = 1800 s).
To calculate the perfect work, apply the formula: A = F * S = F * V * t.
Calculation: A = 50 * 0.1 * 1800 = 9 * 10 ^ 3 J (9 kJ).
Answer: When lifting the taken load, work equal to 9 kJ was performed.