Question

A boy pulls 10lb sledge 30 feet along the horizontal surface t a constant speed, if coefficient of kinetic friction is 0.2 and his pull makes an angle of 45 with the horizintal, what does work he do on the sledge?

Answer 1

Explanation:

1lb is approximately 0.5kg and 1ft is approximately 0.3m. Friction=μN where μ is the coefficient of kinetic friction and N is the normal reaction.

N=mg=(10lb)g=(5kg)g=5g

Friction=5g×0.2=g

work by friction is negative and is equal to the work done by the boy as there is no net change in energy of the object.

work by boy=magnitude of work by friction

approximating g to 10m/s², work by friction = 10×-9= -90J

the work by boy must be |-90j|= 90J

{notice that lb and feet have been approximated to kg and m and the exact values have not been used}