Answer:
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
