In the figure shown below, a 520 a box is placed in front of a spring, with spri

Question

In the figure shown below, a 520 a box is placed in front of a spring, with spring constant 5.5 N/cm, that is compressed a distance Delta x from its natural length. When the box is let go, the spring will push the box and it will slide up the frictionless hill of height H = 1.5 m. At the top of the hill, the ground id not frictionless. The coefficient of kinetic in this area is 0.15 and the coefficient of static friction is 0.25. The box is will travel a distance d = 7.0 cm on the rough surface at the top of the hill before coming to a stop. What is the total work that friction does in bringing the box to a stop? Before entering the area with friction what is the kinetic energy of the box?

Explanation / Answer

1. friction force = mu*mg = 0.25*0.520*9.8 = 459.914 N
W = fs = 459.914*0.07 = 32.19 J

2. KE of the box = work done by friction = 32.19 J
3. PE = mgh = 0.520*9.8*1.5 = 7.644 J
Total E = PE + KE = 39.834 J

4. Total mechanical energy is the same as the top of hill (frictionless surface)
so TE = 39.834 J

5. PE stored in spring = mechanical energy of block = 39.834 J
6. 0.5kx^2 = 39.834
x^2 = 39.834*2/5.5
x = 3.805 m

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