A block of mass m = 2.00 kg is attached to a spring of force constant k = 5.15 102 N/m that lies on a horizontal frictionless surface as shown in the figure below. The block is pulled to a position xi = 5.75 cm to the right of equilibrium and released from rest.
(a) Find the the work required to stretch the spring.
_____ J
(b) Find the speed the block has as it passes through equilibrium.
______ m/s
work=energy in spring=1/2 k x^2
KE=energy in spring
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(a) To find the work required to stretch the spring, we can use the formula for the potential energy stored in a spring:
Potential energy, PE = 1/2 * k * x^2
Where k is the force constant and x is the displacement from equilibrium.
Given:
m = 2.00 kg
k = 5.15 * 10^2 N/m
xi = 5.75 cm = 0.0575 m
We can find the displacement x using:
x = xi - x_equilibrium = 0.0575 m - 0 m = 0.0575 m
Now, we can substitute the values into the potential energy formula:
PE = 1/2 * k * x^2
= 1/2 * (5.15 * 10^2 N/m) * (0.0575 m)^2
Calculating this will give us the answer for the work required to stretch the spring.
(b) To find the speed of the block as it passes through equilibrium, we can use the principle of conservation of mechanical energy. In this case, the total mechanical energy of the system is conserved.
At the initial position, all the energy is in the form of potential energy stored in the spring. When the block passes through the equilibrium position, all of the potential energy is converted to kinetic energy.
Using the law of conservation of mechanical energy:
Potential energy, PE = Kinetic energy, KE
PE = 1/2 * m * v^2 (v is the speed of the block at equilibrium)
Now we can equate the two equations and solve for v:
1/2 * k * x^2 = 1/2 * m * v^2
Substituting the known values:
1/2 * (5.15 * 10^2 N/m) * (0.0575 m)^2 = 1/2 * (2.00 kg) * v^2
Calculating this equation will give us the answer for the speed of the block at equilibrium.
To find the work required to stretch the spring, we can use the equation for the potential energy stored in a spring:
U = (1/2) k x^2
where U is the potential energy, k is the force constant of the spring, and x is the displacement from equilibrium.
In this case, the block is pulled to a position xi = 5.75 cm to the right of equilibrium, so the displacement x = 0.0575 m (converting from cm to m).
Substituting the given values into the equation:
U = (1/2) * (5.15 * 10^2 N/m) * (0.0575 m)^2
Calculating this expression will give us the potential energy stored in the spring, which is equal to the work required to stretch the spring.
For part (b), we can use the principle of conservation of mechanical energy. At the position of equilibrium, all the potential energy stored in the spring will be converted to kinetic energy of the block.
The potential energy stored in the spring at equilibrium is given by:
U = (1/2) k x^2
where U is the potential energy, k is the force constant of the spring, and x is the displacement from equilibrium.
Since the block is at equilibrium, x = 0. Plugging this in:
U = (1/2) k (0)^2
U = 0
This means that all the potential energy is converted to kinetic energy:
(1/2) m v^2 = U
where m is the mass of the block, v is the velocity of the block at equilibrium, and U is the potential energy stored in the spring.
Rearranging this equation to solve for v:
v^2 = (2 * U) / m
Substituting the known values:
v^2 = (2 * U) / (2 kg)
Taking the square root of both sides will give us the speed of the block as it passes through equilibrium.