A 60 kg skier with an initial speed of 12 m/s coasts up a 2.5 m high rise that makes a 35 degree angle with the horizontal. If the coefficient of friction between the skis and the snow is .08, find the speed of the skier when he reaches the top of the rise.

How much work does friction do on the skier?

This is a conservation of energy type of problem!

The work done on friction=mu*mgCos35*2.5/sin35

change of PE=mg*2.5

InitialKE=FinalKE-changePE-frictionwork
find the final speed.

The final formula should read

FinalKE= InitialKe-ChangePE-Wfriction

hope this helps anyone doing this problem for homework!

To solve this problem, we can use the principle of conservation of energy.

First, let's calculate the potential energy (PE) of the skier at the bottom of the rise. The formula for potential energy is:

PE = mgh

where m is the mass of the skier (60 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the rise (2.5 m).

PE = (60 kg) * (9.8 m/s^2) * (2.5 m)
PE = 1470 J

Next, let's find the work done by friction. The work done by friction can be calculated using the formula:

Frictional work = coefficient of friction * normal force * distance

The normal force can be calculated as the weight of the skier since the skier is on a flat surface. The normal force, N, is equal to the gravitational force, which is m * g.

N = (60 kg) * (9.8 m/s^2)
N = 588 N

The distance covered by the skier is the horizontal displacement along the incline. This can be calculated using the formula:

Distance = h / sin(angle)

Distance = 2.5 m / sin(35 degrees)
Distance = 4.35 m

Now, let's calculate the frictional work:

Frictional work = (0.08) * (588 N) * (4.35 m)
Frictional work ≈ 206.736 J

Since we know that the initial kinetic energy (KE) of the skier is equal to the sum of potential energy and kinetic energy at the top:

KE_initial = PE + KE_final

KE_initial = PE + 1/2 * m * (vf)^2

where vf is the final velocity of the skier.

Since the skier starts with an initial speed of 12 m/s and there is no horizontal force acting on the skier, the final velocity will be equal to the initial velocity.

KE_initial = PE + 1/2 * m * (12 m/s)^2

Solving for KE_initial:

KE_initial = 1470 J + 1/2 * (60 kg) * (12 m/s)^2
KE_initial = 1470 J + 4320 J
KE_initial ≈ 5790 J

Now, we can solve for vf:

KE_initial = PE + 1/2 * m * (vf)^2

5790 J = 1470 J + 1/2 * (60 kg) * (vf)^2

Rearranging the equation:

1/2 * (60 kg) * (vf)^2 = 5790 J - 1470 J

1/2 * (60 kg) * (vf)^2 = 4320 J

Simplifying and solving for vf:

(vf)^2 = (4320 J) * (2 / (60 kg))
(vf)^2 ≈ 144 J/kg

vf ≈ sqrt(144 J/kg)
vf ≈ 12 m/s

Therefore, the speed of the skier when he reaches the top of the rise is approximately 12 m/s.