When a 4.20-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.20 cm. (a) If the 4.20-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it? cm (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position

a) 1.27cm b) 1.499Joules

Explanation:

Hooke's law States that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force/load

k is the elastic constant

e is the extension

From the formula, if k = F/e, then

F1/e1 = F2/e2 = k.. (1)

If a mass of 4.2kg causes an extension of 2.20cm, then

F1 = 4.2kg, e1 = 2.20cm

In order to know how far the spring stretches if a 1.50-kg block is hung on it, we will look for the extension e2 caused by the 1.5kg load.

If F2 = 1.5kg, e2 = ?

Substituting the values in equation 1, we will have;

4.2/2.20 = 1.5/e2

Cross multiplying

4.2e2 = 2.20×1.5

4.2e2 = 3.3.

e2 = 4.2/3.3

e2 = 1.27cm

Thus means the spring will stretch by 1.27cm when a load of 1.5kg us hung on it.

b) Work done by an external agent to stretch the same spring 4.00 cm from its unstretched position can be gotten using the expression

W = 1/2Fe

Given e = 4.00cm

F = ?

Before we can get the work done, first we need to know the load that can cause a 4.00cm extension in the spring.

Using F1/e1 = F/e... (3)

Since F1 = 4.2kg, e1 = 2.2cm, e = 4.00cm

Substituting this values into equation 3 to get the load F,

4.2/2.2 = F/4.0

Cross multiplying

2.2F = 4.2×4

2.2F = 16.8

F = 16.8/2.2

F = 7.64kg

If a load of 7.64kg causes an extension of 4cm in the string, work done on the string will be;

W = 1/2Fe

F = mg = 7.64×9.81 = 74.95N

e = 4cm = 0.04m

W = 1/2×74.95×0.04

W = 1.499Joules