A worker pushes a 35.5 kg wheelbarrow 5.00 m along a level surface, exerting a c

Question

A worker pushes a 35.5 kg wheelbarrow 5.00 m along a level surface, exerting a constant horizontal force of 50.0 N. A frictional force of 43.0 N acts on the wheelbarrow in a direction opposite to that of the worker. You must show your work to get credit for this problem 1. a) What is work done by the worker on the wheelbarrow during this motion? Work done by the worker b) What is the work done by friction during this motion? Work done by friction on the wheelbarrow c) What is the net work done on the wheelbarrow during this motion? N:Fd (5433S Net work done on the wheelbarrow d) Assuming the wheelbarrow starts from rest, what is its final speed, i.e. its speed when it has traveled 5.0 m? Speed at 5.0 m- e) What is its change in kinetic energy from the start of its motion to the 5.0 m mark? Change in KE- f) Does wnet-KE?

Explanation / Answer

SOLUTION:

Fnet = 50-43 = 7N

a = Fnet/m = 7/35.5 = 0.197 m/s^2


a) work done by worker = 50*5*cos(0) = 250 J

b) work done by friction = 43*5*cos(180) = -215 J
c) networkdoene = 250-215 = 35 J

d) v^2 - u^2 = 2*a*s

v = sqrt(2*a*s) = sqrt(2*0.197*5) = 1.4 m/s

e)networkdone = change in KE = 35 J

f) Yes

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