Question

An 8.0-9 bullet is shot into a 5.0-kg block, at rest on a frictionless horizontal surface (see the figure). The bullet remains lodged in the block and the bullet-block system moves to the right with speed V. The bullet-block system moves into an ideal massless spring and compresses it by 8.7 cm. The spring constant of the spring is 1200 N/m. V 5.0 k kg el el

If the horizontal surface has a coefficient of kinetic friction μk=0.500, what is the maximum compression of the spring?

Answer 1

Given that:

Spring constant, K = 1200 N/m

Initial compression, x = 8.7 cm = 0.087 m

Energy stored in the spring will be:

E = 0.5 Kx²

E = 0.5 *(1200)*(0.087)²

E = 4.5414 J

According to Law of conservation of momentum, this is the initial kinetic energy.

Now, when friction is there, some energy will be used to do work against friction which is given by:

W = μ_kmg.X

Where X is the compression of the spring/distance covered by mass.

W = (0.5) x (5*9.8) *X

W = 24.5 X

Similarly, energy stored in spring will be:

E' = 0.5 KX²

E' = 0.5 *(1200)*(X)²

E'= 600 X²

Apply law of conservation of energy:

E = E' + W

4.5414 = 600 X² + 24.5 X

600 X² + 24.5 X - 4.5414 = 0

Solving the quadratic equation, we get:

X = 0.06895 m = 6.895 cm --------- (**Answer**)

That is, compression of the spring when there is friction will be : 6.895 cm

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