The Hoist — Two Roads to the Same Watt
A building-site hoist raises a 500 kg pallet 6.0 m up its mast at a steady speed. There are two ways to work out the power it is developing: divide the work by the time, or multiply the force by the speed. One slider sets the speed, and the two columns on the right compute their way to the answer independently. They never disagree — and the reason they cannot is worth more than either calculation.
इस सिमुलेशन का उपयोग कैसे करें
- Steady lifting speed — the single slider, 0.10 to 1.50 m/s, with 0.50 m/s marked on the track. The hoist runs in real time at whatever you set, so the clock is honest.
- Drag anywhere on the scene to walk round the mast.
क्या देखें
- The two power figures agree to the last digit at every speed on the slider.
- Drag from 0.50 to 1.00 m/s: the time halves from 12 s to 6 s, the work stays at 29.4 kJ, and the power doubles. Half the time, double the power, the same job done.
- The force arrow never changes length however fast the hoist runs — at a constant speed the cable pulls with the weight and nothing more. Only the velocity arrow grows.
- Slowing the hoist right down does not reduce the work; it reduces the rate. The pallet still arrives 6.0 m higher, 29.4 kJ better off.
इसके पीछे की भौतिकी
The work done on the load is W = mgh = 500 × 9.8 × 6.0 = 2.94 × 10⁴ J. Route one: the lift takes t = h/v, so P = W/t. Route two: because the cabin rises at a constant speed there is no acceleration, so the cable force is exactly the weight, F = mg = 4900 N, and P = Fv cos θ with θ = 0. Substitute t = h/v into P = mgh/t and the height cancels: P = mgh ÷ (h/v) = mgv = Fv. Same physics, written twice. At 0.50 m/s the lift takes 12 s and both routes give 2450 W ≈ 2.5 kW. Use W/t when a question hands you a height and a time; use Fv when it hands you a speed.