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anonymous
 4 years ago
A boy coasting on a sled has a speed of 15 ft/s when he encounters a 30 degrees hill. The boy sled have a total weight of 60 lb and the coefficient of kinetic friction between sled and hill is .1. What is the boy's acceleration up the hill and how far does he go?
anonymous
 4 years ago
A boy coasting on a sled has a speed of 15 ft/s when he encounters a 30 degrees hill. The boy sled have a total weight of 60 lb and the coefficient of kinetic friction between sled and hill is .1. What is the boy's acceleration up the hill and how far does he go?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can u solve this for me?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1First, draw a diagram. When he is on the slope he has two forces acting on him: gravity \( F_g \) and friction \( F_f \). Create a coordinate frame where the xaxis is parallel to the hill and y is perpendicular to it. The normal force of the hill on the sled is in the y direction. Resolving forces, \[ N = F_g \sin(30) = \frac{1}{2}mg \] and the net force in the y direction is zero. Hence \[ F_f = \mu N = (0.1)\frac{1}{2}mg = 0.05 mg \] and this force is in the negative x direction, decelerating the sled. Now that means the equations of motion in the x direction are  for velocity in the xdirection: \[ v(t) = 15  0.05mgt \]  for dispalcement in the xdirection: \[ x(t) = 15t  \frac{0.05}{2}gt^2 \] Now you should be able to solve your problem. The sled reaches as far as it reaches when the velocity v(t) = 0. Use the value of t you find that from that equation to find the corresponding x(t)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1**correction: the calculation of the normal force above with sin isn't right. It should be \[ N = F_g \cos(30) \] Apologies. You'll need to track that change through the problem.
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