A man pushes a lawnmower 50ft in 5s exerting a force of 30Ib at an angle of 53 degrees with the horizon Find the work done and the average power expended in horsepower?

Answer 1

Work is defined as the product of a force acting through a distance (parallel to it). Power is the rate of work being done per second.

W = F * d
(that notation just means "parallel", and the two are being multiplied).

Since our force isn't totally parallel to his motion already, we have to only use the component of it that IS parallel. That component is "F*cos(53˚)" (assuming he is pushing the mower parallel to "the horizontal").

F = F*cos(53˚) lb
F = 30lb*cos(53˚) lb
F = 18.05 lb
(approx.)

Now that we've found F, multiply it by the distance through which he pushes it.

W = F * d ftlb
F = 18.05 lb
d = 50ft
W = 902.72 ftlb

This is work done. Average power is the work done over the time elapsed:

P = W/Δt
P = 902.72 ftlb / 5s
P = 180.54 ftlb/s

To convert to horsepower, we use the conversion equation

1hp = 550ftlb/s

P = 180.54/550 hp
P = 0.33 hp
(approx.)

Answer 2

The work done is given by the formula: work = force * distance * cos(angle). Plugging in the values, we get work = 30 * 50 * cos(53) = 1200 ft-lb. To find average power, use the formula: power = work / time. Time is given as 5s. Convert the work to horsepower by dividing by 550 (1 horsepower = 550 ft-lb/s), to get 1200 / (5 * 550) = 0.436 hp.