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anyone1979
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[SOLVED] Overcoming friction and then some.
Can anybody help me figure if I have this right?
You push a loaded wheelbarrow of mass 130 kg, exerting a horizontal force of 120 N as you do so. As the wheelbarrow moves, the frictional force acting on it is 85 N.
(a) If you begin from rest, how far do you push the wheelbarrow before reaching a moderate walking speed of 1.3 m/s?
(b) Over this distance, how much work do you do on the wheelbarrow?
(c) How much work does the frictional force do on the wheelbarrow?
(d) What is the final kinetic energy of the wheelbarrow?
a) a = 120/130 = .9m/s^2
1.3^2 = 2(.9)(x)
1.69/1.8 = .94 m
b) W = 120(.94) = 112.8J
c) W = 85(.94)(-1) = -79.9J
d) K = 1/2(130)(1.3^2) = 109.85J
Can anybody help me figure if I have this right?
You push a loaded wheelbarrow of mass 130 kg, exerting a horizontal force of 120 N as you do so. As the wheelbarrow moves, the frictional force acting on it is 85 N.
(a) If you begin from rest, how far do you push the wheelbarrow before reaching a moderate walking speed of 1.3 m/s?
(b) Over this distance, how much work do you do on the wheelbarrow?
(c) How much work does the frictional force do on the wheelbarrow?
(d) What is the final kinetic energy of the wheelbarrow?
a) a = 120/130 = .9m/s^2
1.3^2 = 2(.9)(x)
1.69/1.8 = .94 m
b) W = 120(.94) = 112.8J
c) W = 85(.94)(-1) = -79.9J
d) K = 1/2(130)(1.3^2) = 109.85J