Mary and her sister are playing with a cardboard box on the neighborhood hill. Mary climbs into the box, the total mass of the box with Mary in it is 115 kg. The box starts at rest at the beginning of the incline. The hill is at an incline of 28 degrees with respect to the horizontal.The static and kinectic friction between the box and hill is 0.4 and 0.2 respectively. Assume the box is sliding downhill. a. What is the magnitude of the acceleration of the box? (same as Problem 3b) b. If the hill is 5m tall, what is the Mary's speed when she reaches the bottom? c. At the bottom of the hill, the ground becomes level, but the coefficients of friction do not change. How far will Mary slide before she comes to a stop?

Elements Of Electromagnetics
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Mary and her sister are playing with a cardboard box on the neighborhood hill. Mary climbs into the box, the total mass of the box with Mary in it is 115 kg. The box starts at rest at the beginning of the incline. The hill is at an incline of 28 degrees with respect to the horizontal.The static and kinectic friction between the box and hill is 0.4 and 0.2 respectively. Assume the box is sliding downhill.

a. What is the magnitude of the acceleration of the box? (same as Problem 3b)
b. If the hill is 5m tall, what is the Mary's speed when she reaches the bottom?
c. At the bottom of the hill, the ground becomes level, but the coefficients of
friction do not change. How far will Mary slide before she comes to a stop?

Expert Solution
Step 1

(a)

Draw the free body diagram:

Mechanical Engineering homework question answer, step 1, image 1

Apply Newtons second law in vertical direction direction:

N-mg cosθ=0N=mg cosθ     .....(1)

Apply newtons law in horizontal direction:

mg sinθ-f=mamg sinθ-μkN=mamg sinθ-μkmg cosθ=ma     .....(2)

Enter the required data in equation (2):

mg sinθ-μkmg cosθ=ma  g sinθ-μkg cosθ=a  9.81 m/s2 sin 28°-0.29.81 m/s2 cos28°=a  a=2.88 m/s2

 

 

 

Step 2

(b)

Find the length (s) of the incline:

s=hsin 28°s=5 msin 28°s=10.65 m

Find the velocity at the end of incline as follows:

v2-u2=2asv2-0=22.88 m/s210.65 mv=7.83 m/s

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