A 3.2 kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of this spring/mass sytem is 2.2 J. What is the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.6 kg block?
PE= 1/2 kx^2 but W= kx or x= Weight/k
PE= 1/2 k (Weight/k)^2 reduce it.
9.216
To find the elastic potential energy of the system when the mass is replaced, we need to use the formula for elastic potential energy: PE = 1/2 kx^2, where PE is the elastic potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the elastic potential energy of the system is given as 2.2 J when the mass is 3.2 kg. We can rearrange the formula to solve for k:
PE = 1/2 kx^2
2.2 J = 1/2 k (Weight/k)^2
Next, we need to find the value of x, the displacement from the equilibrium position, in terms of the weight of the object. Using the relationship W = kx, where W is the weight of the object, we can solve for x:
x = W/k
Now, substituting the value of x in terms of W into the equation for potential energy, we get:
2.2 J = 1/2 k (W/k)^2
To find the elastic potential energy of the system when the mass is replaced by 5.6 kg, we can use the same equation but with the new weight. We can rearrange the equation to solve for the new potential energy:
PE_new = 1/2 k (NewWeight/k)^2
Substituting the given values, the new weight is 5.6 kg, which gives:
PE_new = 1/2 k (5.6 kg/k)^2
Simplifying this expression gives the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.6 kg block.