In the figure, a 3.8 kg block is accelerated from rest by a compressed spring of

Question

In the figure, a 3.8 kg block is accelerated from rest by a compressed spring of spring constant 650 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 = 0.293. The frictional force stops the block in distance D = 7.5 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?

https://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/qu_08_dot_55.jpg (Here is the link to figure)

Please di bit yse scientific notation thank you i will make sure to give a great review

Explanation / Answer

Mass of block M = 3.8 kg
Spring constant k = 650 N/m
Coefficient of kinetic friction u = 0.293
Let S = distance travelled by the block after the spring comes to relaxed length.
S = 7.5 m

i) Consider block + bullet as the system under consideration.

Normal force Fn = Mg = 3.8 * 9.8 = 37.24 N
Friction force Ff = u*Fn = 0.293 * 44.1 = 10.91 N
Friction is opposite to displacement.
Work done by friction = -Ff * S = -10.91 * 7.5 = -81.83 J
Increase in thermal energy = - work done by friction = -(-81.83) = 81.83 J

b) Block has maximum KE when the spring's potential energy = 0 i.e. when the spring is in relaxed length.
Let Kmax = maximum KE of the block
When the spring is in relaxed length, then
KE of the system = Kmax
PE of the system = 0

When the block stops, then
KE of the system = 0
PE of the system = 0
Therefore change in mechanical energy=(0+0)-(Kmax +0) = -Kmax
But change in mechanical energy = work done by friction
-Kmax = -81.83 J
Kmax = 81.83J
Ans: 81.83 J

c) Let x = compression in the spring
By conservation of energy
1/2 kx^2 = Kmax
1/2 * 650 * x^2 = 81.83
325 x^2 = 81.83
x = sqrt(81.83/325) = 0.5 m

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