A box of weight mg=9.0 N is pushed up a vertical wall by an applied constant for
ID: 1794343 • Letter: A
Question
A box of weight mg=9.0 N is pushed up a vertical wall by an applied constant force F= 20.0 N at an angle of 30o to the horizontal. The kinetic friction between the wall and the box is constant fk= 0.500N. The box starts from rest and is pushed up a height h=1.50 m. Over this displacment find:
A. The work done by the force on the box ((Hint: what is the correct angle to use between the force and the displacemnt vector: Its 60)?
B.The change in gravitation potential energy of the box-Earth system.
C. The increase in thermal energy of the box and wall due to friction.
Explanation / Answer
A) Workdone by the applied force, W = F*s*cos(theta)
= 20*1.5*cos(90-30)
= 15 J
B) Change in gravitational potential energy = m*g*h
= 9*1.5
= 13.5 J
C) Increase in thermal energy = -workdone by friction
= Fk*h*cos(180)
= 0.5*1.5*(-1)
= -0.75 J
D) Net workdone = increase in kinetci energy
W_Force + F_friction + W_gravity = (1/2)*m*v^2
15 - 0.75 - 13.5 = (1/2)*(9/9.8)*v^2
==> v = 1.28 m/s
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