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A 10 kg block is pushed 25 m up an inclined plane that makes an angle of 35 with the horizontal by a constant?
A 10 kg block is pushed 25 m up an inclined plane that makes an angle of 35 with the horizontal by a constant force of 120N acting parallel to the plane. The coefficient of kinetic friction between the block and the plane is 0.25.
a) how much work is done by the 120N force?
b) what is the increase in the potential energy of the block?
c) what is the work done against friction?
d) what is the increase in the kinetic energy of the block?
e) what is the velocity of the block at the top of the incline?
1 Answer
- civil_av8rLv 78 years agoFavorite Answer
a) W = F*d = 120 N * 25 m = 3000 J
b) Change in PE = m*g*h = 10 kg * 9.81 m/s^2 * 25*sin(35) m = 1410 J
c) W = friction * d
Sum of the force in the perpendicular to the plane direction = 0 (no movement off the plane)
0 = N - m*g*cos(35)
N = m*g*cos(35)
Definition of friction
friction = u*N = u*m*g*cos(35)
friction = 0.25 * 10 kg * 9.81 m/s^2 * cos(35) = 20.1 N
Work = friction * d = 20.1 N * 25 M = 503 J
d) Change in energy = Work
Change in PE + Change in KE = F*d - friction*d
Change in KE = F*d - friction * d - Change in PE = 3000 J - 503 J - 1410 J = 1090 J
e) Assume initil velocity is 0
Change in KE = 1/2*m*v^2
v = sqrt(KE * 2 / m) = sqrt( 1090 J * 2 / 10 kg) = 14.8 m/s