A block of mass 2 kg attached to a spring lies on a flat surface. When the block is pulled 5 cm to the right a force with a magnitude of 300 N is need to hold it in place. What is the work done by the spring when the block moves from x = 10 cm to x = 12 cm. Express your answer in joules.
k=300N/.05m=6000 N/m
work done= PEmax-PEmin
= 1/2 k (.12^2-.10^2)
To find the work done by the spring when the block moves from x = 10 cm to x = 12 cm, we first need to determine the spring constant (k) and the displacement of the block (delta x).
Given information:
- Mass of the block (m) = 2 kg
- Force required to hold the block in place (F) = 300 N
- Initial position (x1) = 10 cm
- Final position (x2) = 12 cm
To find the spring constant (k), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position:
F = -k * x
Rearranging the equation, we get:
k = -F / x
Substituting the given values, we find:
k = -300 N / 0.05 m (converting cm to meters)
k = -6000 N/m
Now that we know the spring constant, we can calculate the displacement of the block:
delta x = x2 - x1
delta x = 12 cm - 10 cm
delta x = 0.02 m (converting cm to meters)
Next, we can use the formula for work done by a spring:
Work = (1/2) * k * (delta x)^2
Substituting the values, we find:
Work = (1/2) * (-6000 N/m) * (0.02 m)^2
Work = -1.2 J
Therefore, the work done by the spring when the block moves from x = 10 cm to x = 12 cm is -1.2 Joules.