Asked by Anonymous
A boy pulls a sled of mass 5.0 kg with a rope that makes an 60.0° angle with respect to the
horizontal surface of a frozen pond. The boy pulls on the rope with a force of 10.0 N; and the
sled moves with constant velocity. What is the coefficient of friction between the sled and the
ice?
horizontal surface of a frozen pond. The boy pulls on the rope with a force of 10.0 N; and the
sled moves with constant velocity. What is the coefficient of friction between the sled and the
ice?
Answers
Answered by
drwls
Constant velocity means that the forces add up to zero.
Do a force balance in the direction of motion. It says:
10.0 cos 60 = friction force = (M g - 10.0 sin 60)*Uk
Solve for Uk, the coefficient of (kinetic) friction.
The reason for the -10.0 sin 60 term is that the upward pull on the rope reduces the normal force of the sled onto the ice. That term gets subtracted from the sled's weight.
Do a force balance in the direction of motion. It says:
10.0 cos 60 = friction force = (M g - 10.0 sin 60)*Uk
Solve for Uk, the coefficient of (kinetic) friction.
The reason for the -10.0 sin 60 term is that the upward pull on the rope reduces the normal force of the sled onto the ice. That term gets subtracted from the sled's weight.
Answered by
asdfghj
.124
Answered by
bigdaddy
.6534
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