A 174g block is launched by compressing a spring of constant k =200N/m a distanc

Question

A 174g block is launched by compressing a spring of constant k=200N/m a distance of 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction ?=0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise, as shown in the figure.(Figure 1)

After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.

Express your answer using two significant figures. s= ? cm

Explanation / Answer

I am going to assume that you mean a 176 gram block. If you mean 176 kg, 1.76 kg, or 17.6 kg, you can follow the same procedure, you just have to plug that value into the equations instead of 0.176 kg.

The elastic potential energy formed in the compressed spring will all convert into kinetic energy for the mass, and then into friction slowing the block. We can skip the kinetic part, and conver the spring energy straight into work done by the friction.

E1 (energy state 1) = elastic energy = 0.5*k*x^2
W (work) = friction across unknown distance = u*N*d
where d is the unknown distance, and N is the normal force.
N = m*g = 0.176 kg * 9.81 m/s^2 = 1.72656 N

You know all the energy in the spring is converted into work done by the friction to slow the block to a stop, so:

A 176 block is launched by compressing a spring of constant 200 a distance of 15 cm T?

E1 = W
0.5*k*x^2 = u*N*d
0.5*200*(0.15^2) = 0.27*1.72656*d ----> Solving for d:
d = 4.827 meters

That indicates that the block slides much further than the 85 cm of frictional surface. That would mean that the block continues on forever until it encounters another frictional surface to stop it.

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