A microwave oven of mass 13.0kg is pushed a distance 7.65m up the sloping surfac

Question

A microwave oven of mass 13.0kg is pushed a distance 7.65m up the sloping surface of a loading ramp inclined at an angle of 38.5 degree above the horizontal, by a constant force F with a magnitude 150N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.200. Part A What is the work done on the oven by the force F ? Take the free fall acceleration to be 9.80m/s2 . Part B What is the work done on the oven by the friction force? Take the free fall acceleration to be 9.80m/s2 . Part C Compute the increase in potential energy for the oven. Take the free fall acceleration to be 9.80m/s2 . Part D Use your answers to parts (A), (B), and (C) to calculate the increase in the oven's kinetic energy. Take the free fall acceleration to be 9.80m/s2 . Express your answer using two significant figures. Part E Use ?F? =ma? to calculate the acceleration of the oven. Take the free fall acceleration to be 9.80m/s2 . Express your answer using two significant figures. Part F Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling a distance 7.65m . Take the free fall acceleration to be 9.80m/s2 . Express your answer using two significant figures. Part G From this, compute the increase in the oven's kinetic energy. Take the free fall acceleration to be 9.80m/s2 . Express your answer using two significant figures.

Explanation / Answer

a) work done by F = F * 7.65 = 150*7.65 = 1147.5 J

b) friction = mu*m*g*cos 38.5 = 0.2*13 *9.8 * cos 38.5 = 19.940856 N
work by fric = friciton * distance = - 19.940856 * 7.65 = - 152.547547 J - negative because the dispalcemetn is opposite tot hhe direction of friction

c) increase in pot. energy = m*g*7.65*sin 38.5 = 13 * 9.8 * 7.65 * sin 38.5 = 606.70899 J

D) increase in ke. = work done by F + frici - increase in pot energy
= 1147.5 - 152.547547 - 606.70899= 388.243463 J

E) net force = m*a

F - friction - mg sin theeta = m*a

150 - 19.940856 - 13*9.8*sin 38.5 = 13 * a

so accleearion a = 3.9039061 m/sec2

F) a = 3.9039061
s = 7.65 m
u = 0

so v^2 = u^2 + 2*a*s
v^2 = 0 + 2*3.9039061 * 7.65

so final velocity = v = 7.7285033 m/sec

G) increase in k,e, = 0.5 * m*v^2 = 0.5 * 13 * 7.7285033^2 = 388.24346 J

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