A child leaves her book bag on a slide. The bag, which is at the top of the slid

Question

A child leaves her book bag on a slide. The bag, which is at the top of the slide, starts from rest and reaches the bottom in 1.73 s. The mass of the book bag is 2.95 kg, the length of the slide is 3.90 m and the angle of incline is 33.5°. (Assume the +x-axis to be parallel to and down the slide. For all values, enter the magnitude only.)

(a) With what acceleration does the bag go down the slide?
_________ m/s2

(b) What is the friction force acting on the bag?
______N

(c) What is the coefficient of kinetic friction between the bag and the slide?

________
(d) What is the speed of the bag when it reaches the bottom of the slide?
_____m/s

Explanation / Answer

Given that

The bag, which is at the top of the slide, starts from rest and reaches the bottom in a time (t) = 1.73 s.

The mass of the book bag is(m) =2.95 kg,

The length of the slide is (s) =3.90 m

And the angle of incline is(theta) = 33.5°.

Acceleration due to gravity (g) =9.81m/s2

a)

what acceleration does the bag go down the slide is given by

s =ut+(1/2)at2

3.90 =0*(1.73)+(1/2)a(1.73)2

a =2*3.90/(1.73)2 =7.8/2.9929

=2.606m/s2

b)

We know that the net force acting is given by

Fnet =mgsintheta-fk

then fk =mgsin(33.5°) -ma

=(2.95)(9.81)sin(33.5°) -(2.95)(2.606m/s2) =15.972-7.6877=8.1043N

c)

The coefficient of kinetic friction between the bag and the slide is given by

fk =umgcos(33.5)

u =fk/mgcos(33.5) =8.1043N/(2.95)(9.81)cos(33.5) =0.3358

d)Tthe speed of the bag when it reaches the bottom of the slide is given by

v =u+at =0+(2.606m/s2)(1.73s) =4.5083m/s

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