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 1.75 s. The mass of the book bag is 2.50 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? 
 

(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

Here ,

a) let the acceleratio of the bag is a

as distance , d = 0.5 at^2

3.90 = 0.5 * a * 1.75^2

a = 2.54 m/s^2

the acceleration of the bag is 2.54 m/s^2

b)

Now , for the bag ,

mg * sin(theta) - f = m*a

2.50 * 9.8 * sin(33.5) - f = 2.50 * 2.54

solving for f

f = 7.17 N

the frictional force is 7.17 N

c)

as f = u * mg * cos(theta)

7.17 = u * 2.5 * 9.8 * cos(33.5 degree)

u = 0.351

the coefficient of friction is 0.351

d)

at the bottom ,

v = sqrt(2 * a * d)

v = sqrt(2 * 2.54 * 3.90)

v = 4.45 m/s

speed of bag at the bottom of the ramp is 4.45 m/s

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