A 2 kg block is pushed by an external force against a spring with spring constant 131 N/m until the spring is compressed by 2.1 m from its uncompressed length ( 0).
The block rests on a horizontal plane that has a coefficient of kinetic friction of 0.74 but is NOT attached to the sprinr 2 kg 2.1 m 2 kg 131 N/m μ 0.74

After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop?
The acceleration due to gravity is 9.8 m/s .
Answer in units of m.

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moya1316 moya1316 · Answer 1 · 2020-01-10T10:22:05+01:00

Answer:

d = 19.92 m

Explanation:

As in this exercise there is friction we must use the relationship between work and energy

W = ΔEm

Look for energy in two points

Initial. Fully compressed spring

Em₀ = [tex]K_{e}[/tex] = ½ k x²

Final. When the block stopped

[tex]Em_{f}[/tex] = 0

Let's look for the work of the rubbing force

W = fr d cos θ

Since rubbing is always contrary to movement, θ = 180

W = - fr d

Let's use Newton's second Law, to find the force of friction

Y Axis

N- w = 0

N = mg

The equation for the force of friction is

fr = μ N

fr = μ mg

We substitute in the work equation

W = - μ m g d

We write the relationship of work and energy

-μ m g d = 0 - ½ k x²

d = ½ k x² / μ m g

Let's calculate

d = ½ 131 2.1 2 / (0.74 2 9.8)

d = 19.92 m