Suppose a 80-kg skier skis 20.0 m along a hill which makes a constant angle with

Question

Suppose a 80-kg skier skis 20.0 m along a hill which makes a constant angle with the vertical. He drops in altitude by 3.0 m during this trip. Use concepts involving conservation of energy to solve the following.

(A) If there were no friction, determine his speed at the end of the 20.0-m distance, assume he starts with a speed of 4.0 m/s.

(B) Now we add friction. The actual speed of the skier at the end of the 20-m distance is 6.20 m/s. Determine the work done by friction.

(C) Determine the force of friction.

In problem 1, you used conservation of energy to eventually find the force of friction. You could also do it using Newton%u2019s laws. (A) Determine the angle the hill makes with the horizontal. (B) Find the acceleration of the skier and the net force on him. (C) Find the component of weight which points down the hill (that is, parallel to the hill, not perpendicular to the surface)%u2018. (D) From your answers to (B) and (C), determine the force of friction. This should agree with the answer you obtained before.

Explanation / Answer

work done by gravity + work done by friction = change in kinetic energy

(A) in this case friction is zero, hence

work done by gravity = change in kinetic energy

m*g*h = 0.5*m*( (final vel )^2   - (initial vel )^2)

or 80*9.8*3=0.5*80*( final vel )^2 - 16)

or final vel = 8.64 m/s

(B) work done by friction = change in kinetic energy - work done by gravity

                                    = 0.5*m*( (final vel )^2   - (initial vel )^2) ) - m*g*h

                                    = -1454.4 J

(C) work done by friction = dist*force of friction

     force of friction = -1454.4 J/ 20 m = -72.72 N (negative sign indicates it opposes motion of skier)

NEWTONS'S LAWS METHOD:

(A) if hill makes theta angle with the horizontal

           sin(theta) = 3/20

       or   theta = asin(0.15) = 8.62 degree

(B) using eqn v^2 - u^2 = 2as ; a = (6.2^2 - 16)/40 = 0.561 m/s/s

        accn of skier = 0.561 m/s/s

(C) component of weight parallel to hill = mgsin(theta) = 80*9.8*0.15 = 117.6 N

(D) m*aacn = component of weight parallel to hill + force of friction;

or force of friction = 80*(0.561) - 117.6 = -72.72 N

thus, the answer comes out to be the same in both cases

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