A microwave oven of mass 13.0 kg is pushed a distance 9.50 m up the sloping surf

Question

A microwave oven of mass 13.0 kg is pushed a distance 9.50 m up the sloping surface of a loading ramp inclined at an angle of 35.9 above the horizontal, by a constant force F with a magnitude 180 N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.260.

Take the free fall acceleration to be 9.80 m/s2 .

a- What is the work done on the oven by the force F ?

b- What is the work done on the oven by the friction force?

c- Compute the increase in potential energy for the oven.

d- Use your answers to parts (A), (B), and (C) to calculate the increase in the oven's kinetic energy.

e- Use F =ma to calculate the acceleration of the oven.

f- Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling a distance 9.50 m .

g- From this, compute the increase in the oven's kinetic energy.

Explanation / Answer

Refer below figure,

a) W= Fd = 180*9.5 = 1710 J

b) Wf = Ff*d = us*mgcos*d = us*Fn*d =0.260*13*9.8*cos35.9*9.5 = - 71.22 J

c) PE = mg*h = mgdsin = 13*9.8*9.5*sin35.9 = 709.7 J

d) KE = Wnet = W+Wf- PE = 1710 J - 71.22 J - 709.7 J = 923 J

e) Fnet = ma

F – mgsin – Ff = ma

F – mgsin – us*mgcos = ma

180 – 13*9.8*sin35.9 – 0.260*13*9.8*cos35.9 = 13*a       =>   a = 6.04 m/s^2

f) vf^2 = vi^2 +2*a*d

vf^2 = 0^2 +2*6.04*9.5      => vf = 10.7 m/s

g) KE = 1/2m(vf^2-vi^2) = ½*13*(10.7^2 -0^2) = 744.12 J

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