Which scenario would require more power?

Which scenario would require more power?

• A. In lifting 100 kg at constant speed of 3 m/s
• B. In lowering the weight at constant speed of 3m/s
• C. In lowering the weight at a velocity of 3 m/s and acceleration of 1 m/s^2
• D. In lifting up a weight of 100 kg at velocity 3 m/s but accelerating upward 0.5 m/s^2

Answer: (D)

Power is the rate at which you do work, and for vertical motion it comes from the force you apply in the direction of the motion times the velocity: P = F · v. If you lift a 100 kg mass at 3 m/s with no acceleration, you must provide a force equal to the weight mg ≈ 100 × 9.81 ≈ 981 N. The power you deliver is then P ≈ 981 N × 3 m/s ≈ 2.94 kW. If you lower at constant speed, the force you apply upward to cancel gravity is still mg, but the velocity is downward, so the force and velocity are opposite and the instantaneous power you provide is negative (you’re absorbing energy rather than delivering it). The magnitude is the same, about 2.94 kW, just with negative sign if you track the direction of work. If you lower with downward acceleration of 1 m/s², the upward force you apply is mg − ma ≈ 100(9.81 − 1) ≈ 881 N, and with the velocity downward at 3 m/s, the power you provide is about −881 × 3 ≈ −2.64 kW (still energy flow but in the opposite sense and smaller magnitude). In the scenario where you lift upward with an acceleration of 0.5 m/s², the force needed is mg + ma ≈ 100(9.81 + 0.5) ≈ 1031 N, and with the upward velocity of 3 m/s, the power delivered is 1031 × 3 ≈ 3.09 kW. So this case requires more power than the others because both the needed force and the direction of motion align, and the extra upward acceleration increases the force beyond the simple weight.

Power is the rate at which you do work, and for vertical motion it comes from the force you apply in the direction of the motion times the velocity: P = F · v.

If you lift a 100 kg mass at 3 m/s with no acceleration, you must provide a force equal to the weight mg ≈ 100 × 9.81 ≈ 981 N. The power you deliver is then P ≈ 981 N × 3 m/s ≈ 2.94 kW.

If you lower at constant speed, the force you apply upward to cancel gravity is still mg, but the velocity is downward, so the force and velocity are opposite and the instantaneous power you provide is negative (you’re absorbing energy rather than delivering it). The magnitude is the same, about 2.94 kW, just with negative sign if you track the direction of work.

If you lower with downward acceleration of 1 m/s², the upward force you apply is mg − ma ≈ 100(9.81 − 1) ≈ 881 N, and with the velocity downward at 3 m/s, the power you provide is about −881 × 3 ≈ −2.64 kW (still energy flow but in the opposite sense and smaller magnitude).

In the scenario where you lift upward with an acceleration of 0.5 m/s², the force needed is mg + ma ≈ 100(9.81 + 0.5) ≈ 1031 N, and with the upward velocity of 3 m/s, the power delivered is 1031 × 3 ≈ 3.09 kW.

So this case requires more power than the others because both the needed force and the direction of motion align, and the extra upward acceleration increases the force beyond the simple weight.