Question

A 1.30-kg object is held 1.10 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is dropped onto the spring. (a) How far does the object compress the spring?

Answer 1

Answer:

0.345m

Explanation:

Let x (m) be the length that the spring is compress. If we take the point where the spring is compressed as a reference point, then the distance from that point to point where the ball is held is x + 1.1 m.

And so the potential energy of the object at the held point is:

$E_p = mgh$

where m = 1.3 kg is the object mass, g = 10m/s2 is the gravitational acceleration and h = x + 1.1 m is the height of the object with respect to the reference point

$E_p = 1.3 * 10 * (x + 1.1) = 13(x + 1.1) = 13x + 14.3 J$

According to the conservation law of energy, this potential energy is converted to spring elastic energy once it's compressed

$E_p = E_k = kx^2/2 = 13x + 14.3$

where k = 315 is the spring constant and x is the compressed length

$315x^2 = 26x + 28.6$

$315x^2 - 26x - 28.6 = 0$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{26 \pm \sqrt{26^2 - 4*(-28.6)*315}}{2*315}$

$x = \frac{26 \pm 191.6}{630}$

x = 0.345 m or x = -0.263 m

Since x can only be positive we will pick the 0.345m