In the figure below, a 3.1 kg block is accelerated from rest by a compressed spr

Question

In the figure below, a 3.1 kg block is accelerated from rest by a compressed spring of spring constant 640 N/m. The block leaves the spring at the spring?s relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction Muk = 0.25. The frictional force stops the block in distance 8.0 m. (a) What is the increase in the thermal kinetic energy of the block floor system? ____________J (b) What was the maximum kinetic energy of the block? ______________ J (c) What was the original compression distance of the spring? ________________ m m

Explanation / Answer

frictional force stops the block in 8 m

Frictional force = uk*N = uk*m*g = .25*3.1*9.8 = 7.6 N

aceeleration = F/m = 7.6/3.1 = 2.45

so now using equation

v2 - u2 = 2as where v(final velocity) = 0

so u2 = 2*2.45*8

u = 6.26

so maximum kinetic energy = .5*m*v2 = 60.76 J (part b)

.5*kx2 = 60.76

x = .435m(part c)

part a = part b = 60.76 J

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