oldunion said:
Sam, whose mass is 66.0kg, takes off down a 55.0m high, 11.0 degree slope on his jet-powered skis. The skis have a thrust of 190N. Sam's speed at the bottom is 38.0 38m/s.
What is the coefficient of kinetic friction on his skees.
I don't understand why they put a slope or velocity in. So what would the base equation be for solving this.
::::::::::#1 & #2 below answer your specific questions. #3 & #4 provide solution guidance & hints.
1) You are given the final velocity so you can determine the component of NET FORCE acting on Sam PARALLEL to ski slope.
2) You are given the slope so you can determine the component of GRAVITATIONAL FORCE acting on Sam PARALLEL to the ski slope.
3) There are 3 Forces acting on Sam in this problem, each with a component Parallel to and Normal to the Ski Slope (in the following,
K is the Kinetic Friction Coefficient):
:::: a) Gravity:
:::::::::: Component Parallel To Slope={mgSin(11 deg)}
:::::::::: Component Normal To Slope={-mgCos(11 deg)}
:::: b) Jet Ski Thrust:
:::::::::: Component Parallel To Slope={190 N}
:::::::::: Component Normal To Slope={0}
:::: c) Ski Slope Surface
:::::::::: Component Parallel To Slope={
-KmgCos(11 deg)}
<----- Kinetic Friction
:::::::::: Component Normal To Slope={mgCos(11 deg)}
4) Procedure:
:::: a) Determine NET FORCE acting on Sam PARALLEL to Slope using Conservation of Energy:
:::::::::: {Total Kinetic Energy} = {NET FORCE || to Slope}*{Distance Along Ski Slope}
:::::::::: (1/2)mv^2 = {NET FORCE || to Slope}*{(55 m)/Sin(11 deg)}
:::::::::: Solve for {NET FORCE || to Slope}
:::: b) Establish Force Balance Equations and SOLVE for Kinetic Friction Coefficient K:
:::::::::: {NET FORCE || to Slope} = {Sum of All PARALLEL Force Components}
:::::::::: 0 = {Sum of All NORMAL Force Components}
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