In the figure, a 3.7 kg block is accelerated from rest by a compressed spring of

Question

In the figure, a 3.7 kg block is accelerated from rest by a compressed spring of spring constant 630 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 k-0.234. The frictional force stops the block in distance D 8.3 m. What are (a) the increase in the thermal energy of the block-floor system, (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring? 000100 No friction (4h) (a) Number (b) Number (c) Number Units Units Units CS

Explanation / Answer

a) Thermal eneergy of block floor system = magnitude of work done by friction
= force of frcition x distance
= (mu mg ) ( D)
= ( 0.234*3.7*9.8)*(8.3) = 70.42 J

b) Maximum kinetic energy of block is just before leaving the spring.
This kinetic energy becomes zero due to -ve work done by friciton. So maximum kinetic energy is also magnitude of work done by friction = 70.42 J

c) When spring is compressed by distance x, it stores spring potential energy = 1/2 k x2
As block leaves spring, it is in relaxed condition. Potential energy of spring is converted to kinetic energy of block in the process. Hence
70.42 = 1/2 kx2
x = (2*70.42 / 630 )1/2 = 0.47 m

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