Question

Holding onto a tow rope moving parallel to a frictionless ski slope, a 61.8 kg skier is pulled up the slope, which is at an angle of 6.8° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 2.47 m/s and (b) v = 2.47 m/s as v increases at a rate of 0.109 m/s2?

Answer 1

To find the magnitude of the force on the skier from the rope, we need to consider the forces acting on the skier along the slope. When the magnitude of the skier's velocity is constant, the magnitude of the force from the rope is equal to the component of the skier's weight parallel to the slope. When the skier's velocity increases, there is an additional acceleration along the slope, and the magnitude of the force from the rope is equal to the sum of the skier's weight component and the additional acceleration.

Step-by-step explanation:

To find the magnitude of the force on the skier from the rope, we need to consider the forces acting on the skier along the slope. We can break down the forces into two components: the gravitational force acting down the slope and the force from the rope pulling up the slope.

(a) When the magnitude of the skier's velocity is constant at 2.47 m/s, the net force acting on the skier is zero along the slope since there is no acceleration. Therefore, the magnitude of the force from the rope, Frope, is equal to the component of the skier's weight parallel to the slope. Frope = m * g * sin(theta), where m is the mass of the skier, g is the acceleration due to gravity, and theta is the angle of the slope with the horizontal.

(b) When the skier's velocity increases at a rate of 0.109 m/s^2, there is an additional acceleration along the slope. In this case, the net force along the slope is equal to the sum of the force from the rope and the component of the skier's weight parallel to the slope. The magnitude of the force from the rope, Frope, is Frope = m * g * sin(theta) + m * a, where a is the additional acceleration along the slope.

Answer 2

a) Frope= 71.7 N

b) Frope=6.7 N

Step-by-step explanation:

In the figure the skier is simulated as an object, "a box".

a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:

∑F=0

Frope=w*sen6.8°

Frope=71.71N

Take into account that w is the weight that is calculated as mass per gravitiy constant:

w=m*g

`w=61.8Kg*9.8(m)/(s^(2) )`

`w=605.64N`

b) In this case the system has an acceleration of 0.109m/s2. Then, we apply Newton's second law of motion:

F=m*a

F=61.8Kg*0.109m/s2

Frope=6.73N