# A 50 kg skier is pulled up a frictionless ski slope that makes an angle of 8 degrees with the horizontal by holding onto a tow rope that moves parallel to the slope. Determine the magnitude of the force of the rope on the skier at an instant when a) the rope is moving with a constant speed of 2 m/s and b) the rope is moving with a constant speed of 2 m/s but that speed is increasing at a rate of .10 m/s^2.

I would need to use F=ma but how would i do this if there is a v involved? I also don't get part b) becasue of the speed that is increasing at a rate of .10 m/s^2

Ren (Lars) Stop posting under multiple names.

Constant speed is zero acceleration. So you are dealing with just force here, the up the slope forces equal the down the slope forces.

Now when there is a changing speed, there is a net acceleration.

Net force= mass*acceleration
When acceleration is zero,
Net force = UPForce -DOWNforce=0
When acceleration is not zero.
UPforce - Downforce= mass*acceleratin

## To solve part a) of the problem, where the rope is moving with a constant speed of 2 m/s, we can assume that there is no acceleration involved. Therefore, the net force acting on the skier is zero.

We can start by calculating the gravitational force acting on the skier. The gravitational force (mg) can be calculated by multiplying the mass (m) of the skier (50 kg) by the acceleration due to gravity (9.8 m/s^2).

Gravitational Force (mg) = (50 kg) * (9.8 m/s^2) = 490 N

Since there is no acceleration, the force of the rope must be equal in magnitude and opposite in direction to the gravitational force acting on the skier. Therefore, the magnitude of the force of the rope on the skier is also 490 N.

For part b) of the problem, where the rope is moving with a constant speed of 2 m/s and increasing at a rate of 0.10 m/s^2, we have a non-zero acceleration. In this case, we need to consider the net force acting on the skier.

The gravitational force acting on the skier is still 490 N, as calculated before.

Now, we need to consider the acceleration of the skier. The acceleration is given as 0.10 m/s^2. Since the rope is parallel to the slope, the acceleration can be decomposed into two components: one parallel to the slope (parallel acceleration) and one perpendicular to the slope (perpendicular acceleration).

The parallel acceleration can be calculated by multiplying the overall acceleration by the sine of the angle of the slope. In this case, the angle is given as 8 degrees.

Parallel Acceleration = (0.10 m/s^2) * sin(8 degrees) = 0.017 m/s^2 (approx.)

The perpendicular acceleration can be calculated by multiplying the overall acceleration by the cosine of the angle of the slope.

Perpendicular Acceleration = (0.10 m/s^2) * cos(8 degrees) = 0.099 m/s^2 (approx.)

Now, we need to calculate the net force acting on the skier.

Net Force = (mass) * (overall acceleration) = (50 kg) * (0.10 m/s^2) = 5 N

To find the magnitude of the force of the rope on the skier, we need to calculate the sum of the gravitational force and the net force.

Magnitude of the Force of the Rope = |Gravitational Force| + |Net Force|

Magnitude of the Force of the Rope = |490 N| + |5 N| = 495 N

Therefore, the magnitude of the force of the rope on the skier at an instant when the rope is moving with a constant speed of 2 m/s and increasing at a rate of 0.10 m/s^2 is 495 N.