A skier moving at 5.71 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping?

She travels 7.56 m before stopping.

Explanation:

Hi there!

According to the work-energy theorem, the magnitude of the work done by a force on a moving object to bring it to stop will be equal to the kinetic energy of the object:

W = KE

Where:

W = work

KE = kinetic energy

In this case, the force that stops the skier is the friction force. Then, the work done by friction will be:

W = Fr · d

Where:

Fr = friction force.

d = traveled distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of kinetic friction.

The forces acting in the vertical direction are the weight of the skier (w, downward) and the normal force (N, upward). Since the skier is not being accelerated in the vertical direction, then the sum of vertical forces is equal to zero:

∑Fy = N - w = 0 ⇒N = w

The weight is calculated as follows:

w = m · g

Where m is the mass of the skier and g is the acceleration due to gravity.

Then, the work done by friction can be expressed as follows:

W = Fr · d

W = N · μ · d

Since N = w = m · g

W = m · g · μ · d

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where v is the speed of the skier.

Appliyng work-energy theorem:

W = KE

m · g · μ · d = 1/2 · m · v²

Solving for d:

d = 1/2 · v² / g · μ

d = 1/2 · (5.71 m/s)² / (9.8 m/s² · 0.220)

d = 7.56 m

She travels 7.56 m before stopping.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

u

explanation: