An object of mass m is at rest at the top of a smooth slope of height h and leng

Question

An object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, micro-kilometer , is small enough that the object will slide down the slope if given a very small push to get it started.


A) Find an expression for the object's speed at the bottom of the slope - expressing your answer in terms of the variables m, micro-kilometers, L, and h, AND appropriate constants.


B) Sam, whose mass is 70kg, stands at top of a 12m-high, 120m-long snow-covered slope. His skis have a coefficient of kinetic friction on snow of 0.07. If he uses his poles to get started, then glides down, what is his speed at the bottom? Expressing your answer to two sig figs and using appropriate units.

Explanation / Answer

let mass be m. given that height=h length=L so if the slope makes angle theta with horizontal,sin(theta)=h/L now gravitational force down the incline=m*g*sin(theta) normal force=m*g*cos(theta) so friction force=k*m*g*cos(theta) so net force down the incline=m*g*(sin(theta)-k*cos(theta)) so acceleration=g*(sin(theta)-k*cos(theta)) velocity after covering distance L=sqrt(2*acceleration*distance)=sqrt(2*g*(sin(theta)-k*cos(theta))*L) where sin(theta)=h/L cos(theta)=sqrt(L^2-h^2)/L b. h=12 L=120 k=0.07 so sin(theta)=h/L=0.1 cos(theta)=sqrt(L^2-h^2)/L=0.995 velocity=8.448 m/s

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