coefficient of kinetic friction between the box and the surface of the incline is 0.182 The initial speed of the box at the bottom of the incline is 3.48 m/s. How far does the box travel along the incline before coming to rest?
2006-10-16 13:30:09 · 3 answers · asked by meggers 3 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*m*3.48^2 = 6.05(m) 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, 12.4°. So the force is 0.182*m*9.8*sin(12.4) = .383(m) N. The path length along which is acts is equal to y/sin(12.4) = 4.65y, so the work done by friction is .383(m)*4.65y = 1.78(m)y 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 = m*9.8y = 9.8(m)y J/m. At the max height, the motion stops, and kinetic energy is zero. So you just need to solve 6.05(m) J = (1.78(m) +9.8(m))y J/m. The solution is y = (6.05(m) J) / (11.58(m) J/m) = .522 m. If you want to know the actual length along the incline, just divide by sin(12.4) to get 2.433 m.
2006-10-16 13:55:16 · answer #1 · answered by resurrection_of_t_o 2 · 0⤊ 0⤋
That's a good question, and I'll bet your physics teacher will be glad to assist you if you're having trouble understanding the principles you need to solve the problem. Science and math can be confusing. But a good teacher can make it easy to understand.
Teachers love to teach. Try it, just ask them a question. Not knowing the answer is fine, as long as you're willing to take responsibility for your learning.
2006-10-16 20:40:29 · answer #2 · answered by Tomis 3 · 0⤊ 2⤋
http://en.wikipedia.org/wiki/Coefficient_of_friction
2006-10-17 00:37:48 · answer #3 · answered by Surya M. 3 · 0⤊ 0⤋