Question

When a 4.32 kg object is hung vertically on a certain light spring that obeys Hooke's Law, the spring stretches 2.92 cm.

Answer 1

(a) 1.01 cm

First of all, we need to find the spring constant of the spring.

The force initially applied to the spring is equal to the weight of the block hanging on it:

$F=mg=(4.32)(9.8) = 42.3 N$

where m = 4.32 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration of gravity.

When this force is applied, the spring stretches by

$x=2.92 cm = 0.0292 m$

We can find the spring constant by using Hooke's law:

$F=kx$

where k is the spring constant. Solving for k,

$k=\frac{F}{x}=\frac{42.3}{0.0292}=1448.6 N/m$

Later, the first object is removed and another object of mass

m' = 1.50 kg

is hung on the spring. The weight of this object is

$F'=m'g=(1.50)(9.8)=14.7 N$

So, if we use Hooke's law again, we can find the new stretching of the spring:

$x'=\frac{F'}{k}=\frac{14.7}{1448.6}=0.0101 m = 1.01 cm$

(b) 1.16 J

The work that must be done on the spring is equal to the elastic potential energy that would be stored in the spring, therefore:

$W=\frac{1}{2}kx^2$

where we have

k = 1448.6 N/m is the spring constant

x = 4.00 cm = 0.04 m is the new stretching

Solving the equation, we find the work that must be done by the external force:

$W=\frac{1}{2}(1448.6)(0.04)^2=1.16 J$