Posted by puff puff on .
A skier of mass 70kg is pulled up a slope by a motor-driven cable. The acceleration of gravity is 9.8m/s^2. How much work is required to pull him 60m up a 30.7º slope (frictionless) at a constant speed of 3 m/s? answer in units of kJ.
In this question, I think I gotta use mgh and 1/2mv^2 to calculate the energy but i don't know how. can anyone help me? thanks in advance!
Forget about kinetic energy in this problem. Since the pulling is being done at constant speed, the kinetic energy does not change and the speed does not matter when computing the work required.
The work energy required equals the change in potential energy,
W = M g H sin 30.7.
(H sin 30.7 is the change in altitude)
M = 70 kg
H = 60 m.
net force = ma (up) = F (applied, up) - mg sin 30
ma = F - mg sin 30
it moves with constant speed >> v = constant, a = dv/dt =0
0 = F - mg sin 30
F = mg sin 30 = 343 N (applied force up)
work done W = Force *displacement cos (theta)
W = F (up) * d (up) cos (0)
W = 343 * 60 = 20580 Joule >>work done by motor
Power (motor) = F (average) * v (constant)
P = 343 *2 = 686 watt