Need Help With A Fairly Easy Physics Problem?

Question

A 20 cm(0.02 m) long spring with spring constant 5000 N/m is placed vertically on the ground and a 10.2 kg block is held 15 cm(0.15 m) above the spring. The block is dropped, hits the spring, and compresses it.

1) What is the height of the spring at the point of maximum compression?

Answer 1

First of all, your conversion of 20 cm to its equivalent meter should be 0.20 m (and not 0.02 m).

To solve this problem, use conservation of energy, i.e., potential energy of block plus the work done by the block on the spring must be equal to the energy absorbed by the spring.

Potential Energy of block = PE = mgh

where

m = 10.2 kg (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

h = 0.15 (given)

Work done by block on the spring = mgx

Also,

The energy absorbed by the spring,

W = 1/2(kx^2)

where

k = spring constant = 5000 N/m (given)

x = compressed length of spring

Substituting appropriate values,

(10.2)(9.8)(0.15) + (10.2)(9.8)x = (1/2)(5000)(x^2)

2500x^2 - 99.6x - 14.994 = 0

Simplifying,

1250x^2 - 49.8x - 7.497 = 0

Using the quadratic equation, solve for "x",

x = 0.10 m

Therefore, the length of the spring at the point of maximum compression is = 0.2 - 0.1 = 0.1 m = 10 cm.

Answer 2

You solve this using conservation of energy. When the block is held above the spring it has gravitational potential energy and when the spring is compressed it has the potential energy of its compression.

m = mass of the block

g = acceleration of gravity

Grav potential energy = G = H*mg

Where H is the difference in the altitudes of the block. it starts out 0.15 meters above the spring and ends up x meters below where the top of the spring was before the block was dropped.  So:

G = H*mg = (0.15 + x)mg

Compressed energy of the spring = (1/2)kx^2

(0.15 + x)mg = (1/2)kx^2

(k/2)x^2 - (mg)x - 0.15mg = 0

This is just a quadratic: ax^2 + bx + c = 0 with:

a = k/2 = 5000/2 = 2500

b = -mg = -10.2*9.8 = 99.96

c = -0.15mg = -14.994

Use quadratic formula:

x = [-b +/- SQRT(b^2 - 4ac)] / (2a)

x = [99.96 +/- 399.91] / (5000)

x = 0.1 meters = 10 cm

Check: Look at the compression at equilibrium:

hk = mg

5000h = 10.2(9.8

h = 0.02 meters

And this should be smaller than the when the spring is at the point of maximum compression and it is.

The height of the spring at maximum compression is:

height = 20 cm - x = 20 cm - 10 cm

height = 10 cm