????? A 4 kg block?

Question

slides on a horizontal frictionless surface with a speed of 1.4 m/s. It is brought momentarily to rest when it hits a bumper that compresses a spring. Acceleration of gravity is 9.8 m/s^2.

How much is the spring compressed if its force constant is 8 kN/m? Answer in units of m.

Answer 1

As the block is slowed by compressing the spring, he block's Kinetic Energy (KE) is transfered into the spring's Potentian Energy (PE). The change in the block's KE equals the change in the spring's PE equals the work done on the spring.

The work done by compressing a spring of spring constant k a distance of x meters is given as,
W = 1/2 kx^2

The blocks original Kinetic Energy is given as,
KE = 1/2 mv^2
where m is the block's mass, and v is the blocks speed prior to collision with the spring.

Set the KE of the block equal to the Word done on the spring and solve for x,
1/2 mv^2 = 1/2 kx^2
x^2 = (mv^2) / k
x = sqrt ((mv^2) / k)
You know the valus of m, v, and k, ..., simplu plug in and obtain your answer.