A horse pulls a cart with a force of 40 lb at an angle of 300 above the horizontal and moves along at a speed of 60.0 mi/h. (a) How much work does the force do in 10 min? (b) What is the average power (in horsepower) of the force?
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Solution
The horse pulls with a force →F.
As the cart moves through a displacement
→d, the work done by
→F is W=→F.→d=Fdcosϕ, where
ϕ is the angle between →F.→d.
(a) In 10 min the cart moves
d=vΔt=(6.0mih)
(5280ft/mi60min/h)(10min)=5280ft. so that ,
W=Fdcosϕ=(40lb)(5280ft)cos300=1.8×105ft.lb.
(b) The average power is given as.
With Δt=10min=600s, we obtain
Pavg=WΔt
=1.8×105ft.lb600s=305ft.lb/s,
which (using the conversion factor 1 hp = 550 ft. lb/s found on the inside back cover of the text) converts to Pavg=0.55hp.