A 0500kg block attached to a spring with length 060 m and fo

A 0.500-kg block attached to a spring with length 0.60 m and force constant 40.0 N/m is at rest with the back of the block at point A on a frictionless, horizontal air table (see figure below). The mass of the spring is negligible. You pull the block to the right along the surface by pulling with a constant F = 21.0-N horizontal force.

(a) What is the block\'s speed when the back of the block reaches point B, which is d = 0.22 m to the right of pointA?

(b) When the back of the block reaches point B, you let go of the block. In the subsequent motion, how close does the block get to the wall where the left end of the spring is attached?

The answer to A is 3.82 M/S

The answer to B is .119

I want to know how to get these answers. Thank you!

Solution

According to the given problem,

A) Using the work-energy theorem

Work done = (F x d) = KE (1/2m v²) + Elastic P.E(1/2kx²)

(21N x 0.22m) = (0.5*0.5kg *v²) + (0.5 x 40N/m x 0.22²m)

4.62 = 0.25v² + 0.968

v² = 3.652 / 0.25

v = 14.608

v = 3.82 m/s

B) Energy in system at release = Work done = 21N x 0.22m = 4.62 J

Energy in system at max compression [d] when KE = 0, = 4.62 J = strain energy in compressed spring

1/2*k*d² = 4.62J

d² = 2*4.62/40

d = 0.231

d = 0.48 m compression distance

Distance from wall = 0.60m (spring length) - d (0.48m)

Answer : 0.119m