A 66.0-kg skier starts from rest andslides down a 33.0-m frictionless slopethat

Question

A 66.0-kg skier starts from rest andslides down a 33.0-m frictionless slopethat is inclined at an angle of 15.0° with the horizontal.Ignore air resistance.
If the slope is not frictionless so that the skier has a finalvelocity of 4 m/s, calculate the work doneby gravity, the work done by the normal force, the work done byfriction, the force of friction (assuming it is constant), and thecoefficient of kinetic friction. > How much work is being done by normal force in kJ? > How much work is being done by friction in kJ?
A 66.0-kg skier starts from rest andslides down a 33.0-m frictionless slopethat is inclined at an angle of 15.0° with the horizontal.Ignore air resistance.
If the slope is not frictionless so that the skier has a finalvelocity of 4 m/s, calculate the work doneby gravity, the work done by the normal force, the work done byfriction, the force of friction (assuming it is constant), and thecoefficient of kinetic friction. > How much work is being done by normal force in kJ? > How much work is being done by friction in kJ?

Explanation / Answer

Work done by grativy. W1=66*33*sin15*9,8=5524(J). Change in kinetic energy K=mv^2/2=528(J). so work done by friction 5524-528=5000(J). Work done by normal force = zero. Coefficient. *m*g*cos15*33=5000 so =0,2425

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