S monkey is strapped to a sled and both are given an initial speed of 2.0 m/s up a 24.0° inclined track. The combined mass of monkey and sled is 14 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move?
2006-10-16 08:06:46 · 3 answers · asked by activegirl 1 in Science & Mathematics ➔ Physics
You can do this problem with conservation of energy. I'm going to assume we're solving for the vertical distance, y, rather than the actual path length along the include. Now, the initial kinetic energy is 0.5*m*v^2 = 0.5*14*2^2 = 28 J. The force of friction is constant, and is equal to umg*sin(theta), where u is the coefficient of friction, .2, and theta is the angle, 24°. So the force is 0.2*14*9.8*sin(24) = 11.2 N. The path length along which is acts is equal to y/sin(24) = 2.5y, so the work done by friction is 11.2*2.5y = 27.44y J/m. (When you multiply that by a distance in meters, you'll get a value in joules.) Also, the gravitational potential energy is mgy = 14*9.8y = 137.2y J/m. At the max height, the motion stops, and kinetic energy is zero. So you just need to solve 28 J = (27.44 +137.2)y J/m. The solution is y = (14 J) / (165 J/m) = 0.17 m = 17 cm. If you want to know the actual length along the incline, just divide by sin(24) to get 41.8 cm.
2006-10-16 08:20:26 · answer #1 · answered by DavidK93 7 · 1⤊ 0⤋
The sled has kinetic skill at the same time as it began the journey.Of this in part replaced into lost in friction & in part switched over to skill skill by technique of distinctive feature of its position above floor. research those 2
2016-12-04 21:44:16 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Total Force = gravitational force + friction force
Fg = m*g*Sin[24] =66.5 N
Ff = uf*m*g = (0.2)*14 kg*9.8 = 27.44 N
Total Force = 66.5 N + 27.44 N = 93.96 N
2006-10-16 08:16:30 · answer #3 · answered by resurrection_of_t_o 2 · 0⤊ 1⤋