A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It takes 3.6 J of work to compress the spring by 0.13 m. If the spring is compressed, and the mass is released from rest, it experiences a maximum acceleration of Find the value of (a) the spring constant and (b) the mass. Giancoli, Douglas C.. Physics (p. 323). Pearson Education. Kindle Edition.

A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It takes 3.6 J of work to compress the spring by 0.13 m. If the spring is compressed, and the mass is released from rest, it experiences a maximum acceleration of 12m/s². Find the value of

(a) the spring constant and

(b) the mass.

Giancoli, Douglas C.. Physics (p. 323). Pearson Education. Kindle Edition.

(a) 426.04N/m

(b) 4.62kg

(a) According to Hooke's law, work done is stored as the potential energy (E) due to the compression of the spring is given by;

E = x k x e² ----------------------(i)

Where;

k = the spring constant of the spring

e = the compression

According to the question;

e = 0.13m

E = 3.6J

Substitute these values into equation (i) as follows;

3.6 = x k x 0.13²

3.6 = x k x 0.0169

3.6 = k x 0.00845

Now solve for k;

k = 3.6 / 0.00845

k = 426.04 N/m

Therefore, the spring constant is 426.04N/m

(b) When the spring is compressed and the mass is released from rest, the same force (F) on the spring will cause the mass to accelerate. i.e

F = k x e        -----------Force on spring [From Hooke's law]

F = m x a        ----------Force causing the acceleration(a)    [From Newton's law]

Combine the two equations as follows;

k x e = m x a             -------------------(ii)

Where a = 12m/s², substitute the other necessary values into equation (ii) as follows;

426.04 x 0.13 = m x 12

55.385 = 12m

Solve for m as follows;

m =

m = 4.62kg

Therefore, the mass is 4.62kg