A physics professor is pushed up a ramp inclined upward at 30.0 above the horizo
ID: 1773067 • Letter: A
Question
A physics professor is pushed up a ramp inclined upward at 30.0 above the horizontal as she sits in her desk chair, which slides on frictionless rollers. The combined mass of the professor and her chair is 85.0 kg. She is pushed 2.50 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor’s speed at the bottom of the ramp is 2.00 m/s. (a) Use workenergy theorem to find her speed at the top of the ramp. (b) What would her speed be if there was constant force of friction between the chair and the incline, with the coefficient of friction k = 0.200?
This question has already been answered but neglected to show exactly how to find the Net Work done on the object.
Explanation / Answer
force applied F = 600 N
component of gravitational force Fg = - M*g*sintheta ( down the incline)
net force Fnet = F +Fg
net work done = F*s*costheta = (F+Fg)*s*cos0 = (F+Fg)*s
from work energy theorem net work = change in kinetic energy
(F + Fg)*s = (1/2)*m*(vf^2 - vi^2)
(600 - (85*9.8*sin30))*2.5 = (1/2)*85*(vf^2-2^2)
vf = 3.85 m/s <<<------ANSWER
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force applied F = 600 N
component of gravitational force Fg = - M*g*sintheta ( down the incline)
frictional force fk = -uk*n = -uk*M*g*costheta
net force Fnet = F +Fg + fk
net work done = F*s*costheta = (F+Fg+fk)*s*cos0 = (F+Fg)*s
from work energy theorem net work = change in kinetic energy
(F + Fg + fk)*s = (1/2)*m*(vf^2 - vi^2)
(600 - (85*9.8*sin30) - (0.2*85*9.8*cos30))*2.5 = (1/2)*85*(vf^2-2^2)
vf = 2.51 m/s <<<------ANSWER
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