Aprofessor has to haul a cart up a ramp. the ramp has an angle of about 10 degrees and is about 5 meters long. his initial speed at the bottom of the ramp is 4.5 m/s, and the cart has a mass of about 80 kg. how hard does he have to push on the cart so that at the top of the ramp, its speed has not dropped below 3.6 m/s? neglecting friction, what is the magnitude of the minimum force he has to exert on the cart?

77.95 N

Explanation:

work done by force WF = F×L

work done by gravity Wg = -m×g×L×sinθ

initial kinetic energy Ki = (1/2)×m×vi^2

final kinetic energy Kf = (1/2)×m×vf^2

from work energy relation

net work = change in KE

W_tot = K_f - K_i

WF + Wg = K_f - K_i

F×L - mg×L×sinθ = (1/2)×m×(v_f^2 - v_i^2)

F×5 - 80×9.81×5×sin10 = (1/2)×80×(3.6^2-4.5^2)

force F = 77.95 N

answer; equivalent to a one(1) horsepower motor;

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