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 1.61 s. The mass of the book bag is 2.20 kg, the length of the slide is 3.30 m and the angle of incline is 29.5°. (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/

Answer 1

Answer:

Part a)

$a = 2.55 m/s^2$

Part b)

$F_f = 5.02 N$

Part c)

$\mu = 0.27$

Part d)

$v_f = 4.1 m/s$

Explanation:

Part a)

As we know that the length of the slide is L = 3.30 m

time = 1.61 s

so we can use kinematics here to find the acceleration of the bag

$L = v_i t + \frac{1}{2}at^2$

$3.30 = 0 + \frac{1}{2}a(1.61)^2$

$a = 2.55 m/s^2$

Part b)

As we know by force equation

$mg sin\theta - F_f = ma$

$(2.20)(9.81)sin29.5 - F_f = (2.20)(2.55)$

so we have

$F_f = 5.02 N$

Part c)

Now we know that perpendicular to the plane we will have

$F_n = mg cos\theta$

$F_n = (2.20)(9.81) cos29.5$

$F_n = 18.78 N$

now we know that

$F_f = \mu F_n$

$5.02 = \mu (18.78)$

$\mu = 0.27$

Part d)

When the bag will reach at the bottom then the final speed is given as

$v_f^2 - v_i^2 = 2 a L$

so we have

$v_f^2 - 0 = 2(2.55)(3.30)$

$v_f = 4.1 m/s$