A spring fixed at one end is compressed from its relaxed position by a distance of 0.20 m. See the graph of the applied external force, Fx, versus the position, x, of the spring.
(a) Find the work done by the external force in compressing the spring 0.17 m starting from its relaxed position.
You have to read from the graph what force (Fx) corresponds to what distance (position, x) of the spring.
The work done is the area of the triangle between x=0 and x=0.17 m.
To find the work done by the external force in compressing the spring, we need to use the formula for work:
Work = Force x Distance
In this case, the force is the external force applied and the distance is the displacement of the spring.
From the graph, we can see that the applied external force increases linearly as the position of the spring increases. This means that the force is directly proportional to the position: Fx ∝ x.
To find the equation that relates the external force (Fx) and the position of the spring (x), we can use the formula of a straight line:
y = mx + c
From the graph, we can see that the line passes through the origin, which means the constant term (c) is zero.
So, the equation relating the external force (Fx) and the position of the spring (x) can be written as:
Fx = kx
where k is the constant of proportionality.
To find the value of k, we can use the given information. When the spring is compressed by a distance of 0.20 m, the applied external force is 6.0 N. So we can set up the following equation:
6.0 N = k(0.20 m)
Solving for k:
k = 6.0 N / 0.20 m = 30 N/m
Now, we have the equation:
Fx = 30x
To find the work done in compressing the spring by 0.17 m, we substitute x = 0.17 m into the equation:
Work = Force x Distance = Fx * x
Work = 30 * 0.17 = 5.1 Joules
Therefore, the work done by the external force in compressing the spring by 0.17 m is 5.1 Joules.