A force of 6 pounds is required to hold a spring stretched 0.2 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?
could you please explain how. thanks
To find the work done in stretching the spring, we need to use Hooke's law, which states that the force required to stretch or compress a spring is directly proportional to the displacement from its natural length.
In this case, the force required to hold the spring stretched 0.2 feet beyond its natural length is given as 6 pounds. This tells us that for each 0.2 feet of displacement, the force required is the same, which is 6 pounds.
Now, we need to determine the force required to stretch the spring from its natural length to 0.6 feet beyond its natural length. Since each 0.2 feet displacement requires a force of 6 pounds, we can calculate the force required for 0.4 feet displacement (0.6 - 0.2).
Force = (Displacement / 0.2) * 6 pounds
Force = (0.4 / 0.2) * 6 pounds
Force = 2 * 6 pounds
Force = 12 pounds
So, the force required to stretch the spring from its natural length to 0.6 feet beyond its natural length is 12 pounds.
To find the work done, we need to multiply the force by the displacement. In this case, the displacement is 0.4 feet (0.6 - 0.2).
Work = Force * Displacement
Work = 12 pounds * 0.4 feet
Work = 4.8 foot-pounds
Therefore, the work done in stretching the spring from its natural length to 0.6 feet beyond its natural length is 4.8 foot-pounds.