A dockside winch reels a fishing boat in with a constant 800 N force, parallel to the motion, at a steady 2 m/s.
P = Fv says the winch delivers 1600 W. Why should a force times a speed make a power?
Run one stretch of the line, and count the work and the seconds.
In 5 s, the fishing boat covers 10 m at its steady 2 m/s.
The work done over that stretch: 800 × 10 = 8000. The winch does 8000 J in 5 s.
Power is work per time: 8000 ÷ 5 = 1600. The winch delivers 1600 W.
Now run the same count with symbols: in a time t, the force acts through a distance d.
The work done is W = F × d, so the power is P = F × d ÷ t.
The piece d ÷ t is the distance covered per time: the speed v.
Check it on the numbers: 800 × 2 = 1600. The same 1600 W, with no stretch named.
In the count P = W/t with W = F × d, what is d ÷ t?
A tugboat doubles its steady speed while its engine applies the same constant force, parallel to the motion. Explain why the power delivered doubles.
Each second, the tug's force acts through the distance the tug covers in that second.
At double the speed, the tug covers double the distance each second.
The same force through double the distance does double the work each second.
Work per second is power, so the power doubles.
A constant 400 N force, parallel to the motion, acts on a cart moving at a steady 3 m/s. Flor says: 'The force does 1200 J of work each second.' Is she correct?
A groundskeeper applies a constant 90 N force to a mower, parallel to the motion. The mower covers 4 m in 2 s at a steady speed. Find the power two ways.
First way, work per time:
90 × 4 = 360. The work done is 360 J.
360 ÷ 2 = 180. The power is 180 W.
Second way, force times speed:
4 ÷ 2 = 2. The steady speed is 2 m/s.
90 × 2 = 180. The power is 180 W.
Both counts give 180 W. P = Fv is the work-per-time count written in symbols.
A constant 240 N force, parallel to the motion, moves a rowboat 15 m in 5 s at a steady speed. Enter the power delivered, in watts.
A delivery truck's engine applies the same constant driving force at a steady 20 m/s as it did at a steady 10 m/s. Why does it deliver more power at 20 m/s?
A constant 300 N force, parallel to the motion, moves a raft 60 m in 12 s at a steady speed. Enter the power delivered, in watts.
In P = Fv, the product F × v equals which quantity?