Question

A student of mass 65.4 kg, starting at rest, slides down a slide 18.2 m long, tilted at an angle of 32.1° with respect to the horizontal. If the coefficient of kinetic friction between the student and the slide is 0.133, find the force of kinetic friction, the acceleration, and the speed she is traveling when she reaches the bottom of the slide. (Enter the magnitudes.)

HINT

(a)

the force of kinetic friction (in N)

N

(b)

the acceleration (in m/s^{2})

m/s^{2}

(c)

the speed she is traveling (in m/s)

m/s

Answer #1

There are three forces acting on the student: mg is the
gravitational force of the Earth, N is the normal force of the
slide on the student, F_{k}is the force of kinetic
friction. The free body diagram of the student is shown below:

The gravitational force is resolved into components parallel and perpendicular to the slide. The student is not allowed to move in the direction perpendicular to the slide, the net force on the student in this direction is zero,

The force of kinetic friction on the student is

Substituting values

**(b)**

The net force in the direction along the slide accelerates the student down the slide. Let a be the acceleration of the student. By Newton's second law of motion, the acceleration of the student is given by

Substituting values

**(c)**

The acceleration of the student is constant down the slide. The motion of the student is governed by the kinematics equations for uniformly accelerated motion.

The student starts from rest, initial velocity of the student is zero u=0.

The distance traveled by the student is d=18.2m

The final velocity is given by

Substituting values

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