As pictured below, a 10 kg block, initially at rest, slides down a frictionless

Question

As pictured below, a 10 kg block, initially at rest, slides down a frictionless slope that is 3.0 m tall. It then travels along a flat rough section that is 6 m long. The block finally collides with a spring of spring constant 2250 N/m and compresses the spring 0.30 m.

13. What is the coefficient of kinetic friction between the block and the surface during the rough section?

a) 0.33 b) 0.66 c) 0.15 d)1.96 e) 0.20 f) 0.50

14. What is the speed of the block right before it hits the spring?
a) 7.7 m/s b ) 2 m/s c) 23 m/s d) 4.5 m/s e)2.4 m/s f) 2.8 m/s

15. If there were NO FRICTION during the entire trip, how much would the spring compress? a) 0.30 m b) 0.23 m c) 0.51 m d) 0.12 m e) 0.66 m f) 0.72 m

Explanation / Answer

Since the slope is frictionless its shape doesn't matter. The potential energy at the top "mgh" will equal the KE as it starts along the rough patch. So one eq. is;
mgh = K1

When it slides, the work done by friction "-fs = -umgs" will cause its KE to drop to "K2' just before it hits the spring. So another eq is;
-umgs = K2 - K1

Finally the PE stored in the compressed spring "(1/2)kx^2" will equal the "K2" . So your last eq is;
K2 = (1/2)kx^2

Now just solve for "u" in the middle eq, since the first & last eqs will allow you to find K1 & K2.

If you want to use the work/energy theorem direct, you note that you start with only potential energy U1 = mgh and you end with only potential energy, U2 = (1/2)kx^2 . The difference equals the work done along the patch;
-umgs = U2 - U1

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