A force of 4 pounds is required to hold a spring stretched 0.1 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?
To find the work done in stretching the spring from its natural length to 0.6 feet beyond its natural length, we can use the equation for work:
Work = Force × Distance
Given:
Force required to hold the spring stretched 0.1 feet beyond its natural length = 4 pounds
We need to find:
Work done in stretching the spring from its natural length to 0.6 feet beyond its natural length.
First, let's find the force constant "k" of the spring. The force constant relates the force required to the displacement.
Hooke's Law states:
Force = -k × displacement
Given:
Force required to hold the spring stretched 0.1 feet beyond its natural length = 4 pounds
Displacement = 0.1 feet
Rearranging the equation:
k = -Force / displacement
Substituting the given values:
k = -4 pounds / 0.1 feet
k = -40 pounds/feet
Now, let's calculate the work done in stretching the spring from its natural length to 0.6 feet beyond its natural length.
Given:
Initial displacement (natural length) = 0 feet
Final displacement (0.6 feet beyond natural length) = 0.6 feet
Work = 1/2 × k × (final displacement)^2 - 1/2 × k × (initial displacement)^2
Substituting the values:
Work = 1/2 × (-40 pounds/feet) × (0.6 feet)^2 - 1/2 × (-40 pounds/feet) × (0 feet)^2
Work = 1/2 × (-40 pounds/feet) × (0.6 feet)^2 - 1/2 × (-40 pounds/feet) × 0
Work = 1/2 × (-40 pounds/feet) × 0.36 square feet
Work = -0.72 foot-pounds
Therefore, the work done in stretching the spring from its natural length to 0.6 feet beyond its natural length is -0.72 foot-pounds.
To find the work done in stretching the spring from its natural length to 0.6 feet beyond its natural length, we need to calculate the change in potential energy.
The potential energy stored in a stretched or compressed spring is given by the formula:
Potential energy = 0.5 * k * x^2
where:
k is the spring constant (a measure of the stiffness of the spring)
x is the displacement from the equilibrium position
Here, we're given that a force of 4 pounds is required to hold the spring stretched 0.1 feet beyond its natural length. Since force (F) is related to the spring constant (k) and displacement (x) by the equation F = k * x, we can find the value of k.
4 pounds = k * 0.1 feet
Solving for k, we get:
k = 4 pounds / 0.1 feet
Now, we can substitute the value of k into the potential energy formula and calculate the energy difference when the displacement changes from 0.1 feet to 0.6 feet.
Potential energy at 0.1 feet = 0.5 * k * (0.1)^2
Potential energy at 0.6 feet = 0.5 * k * (0.6)^2
The work done in stretching the spring from its natural length to 0.6 feet beyond its natural length is the difference between these two potential energies, calculated as follows:
Work = Potential energy at 0.6 feet - Potential energy at 0.1 feet
Now, let's plug in the values:
k = 4 pounds / 0.1 feet
Potential energy at 0.1 feet = 0.5 * k * (0.1)^2
Potential energy at 0.6 feet = 0.5 * k * (0.6)^2
Finally, we can calculate the work done.
work=INT force*dx from 0 to .6
= int(k x dx)= 1/2 k x^2 over limits
where k= 4lbs/.1f= 20 lbs/ft
work= 1/2 k x^2 from zero to x=.6
work= 1/2 (20lbs/ft)*.6^2 ft^2 - 0
= 36 ft-lbs work