A 540 g block is released from rest at height h0 above a vertical spring with spring constant k = 320 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 19.8 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 3h0 above the spring, what would be the maximum compression of the spring?
distance block moves down = ho + .198
loss of potential energy by block =
.540 ( 9.81 ) (ho+.198)
gain of potential energy by spring = (1/2)(320)(.198). This is energy gained by spring and lost by block.
so
.540(9.81)(ho+.198) = (1/2)(320)(.198)
solve for ho
then do it for 3 ho
To solve this problem, we will use conservation of energy principles and Hooke's Law.
Let's start by answering parts (a) and (b) of the question:
(a) We need to find the work done by the block on the spring. Work is defined as the change in mechanical energy, which includes both kinetic energy (KE) and potential energy (PE).
Initially, the block only has gravitational potential energy, given by PE = mgh0, where m is the mass of the block (540 g = 0.54 kg) and g is the acceleration due to gravity (9.8 m/s^2).
When the block compresses the spring, it loses some potential energy and gains elastic potential energy stored in the spring. At the maximum compression, all the initial potential energy is converted into elastic potential energy.
The elastic potential energy stored in a spring is given by PE = (1/2)kx^2, where k is the spring constant and x is the compression or extension of the spring.
Using the given information, we can determine the work done by the block on the spring:
Work = Change in potential energy
Work = PE_initial - PE_final
Work = mgh0 - (1/2)kx^2
Substituting the known values:
Work = (0.54 kg)(9.8 m/s^2)(h0) - (1/2)(320 N/m)(0.198 m)^2
(b) The work done by the spring on the block is equal in magnitude but opposite in sign to the work done by the block on the spring. Therefore, the work done by the spring on the block is -Work.
Now let's move on to part (c) of the question:
(c) To find the value of h0, we can use the work-energy principle. Conservation of energy tells us that the loss of potential energy (PE) equals the gain in kinetic energy (KE) when there is no non-conservative work involved (such as friction):
PE_initial - PE_final = KE_initial - KE_final
At the top of the motion, when the block is released, the final kinetic energy is zero (since it momentarily stops) and the final potential energy is zero.
Therefore, we have:
mgh0 - (1/2)kx^2 = 0
Solving for h0:
h0 = (1/2)kx^2/mg
Using the given values:
h0 = (1/2)(320 N/m)(0.198 m)^2 / (0.54 kg)(9.8 m/s^2)
Finally, let's address part (d) of the question:
(d) If the block were released from height 3h0 above the spring, we need to find the maximum compression of the spring.
Following the same principles, the initial potential energy is now 3 times higher, and we can calculate the new maximum compression (x) using the work-energy principle:
mgh1 - (1/2)kx^2 = 0
Solving for x:
x = √(2(mgh1) / k)
Substituting the known values:
x = √(2(0.54 kg)(9.8 m/s^2)(3h0) / 320 N/m)
Now that you have the equations and the methodology, you can calculate the work done and the values of h0 and x using the provided values in the problem.
Let's break down the problem step-by-step:
(a) To find the work done by the block on the spring, we can use the formula:
Work = (1/2) * k * (Δx)^2
where k is the spring constant and Δx is the displacement of the block from its original position.
Given that the spring constant, k, is 320 N/m and the displacement, Δx, is 19.8 cm (0.198 m), we can substitute these values into the formula:
Work = (1/2) * 320 N/m * (0.198 m)^2
= 6.37 J
Therefore, the work done by the block on the spring is 6.37 J.
(b) The work done by the spring on the block is equal in magnitude but opposite in sign to the work done by the block on the spring. So, the work done by the spring on the block is -6.37 J.
(c) To find the initial height, h0, we can use the principle of conservation of mechanical energy. At the maximum compression of the spring, all the potential energy at the initial height is converted into the elastic potential energy stored in the spring.
The potential energy at height h0 is given by:
Potential Energy = m * g * h0
where m is the mass of the block (540 g = 0.54 kg), g is the acceleration due to gravity (9.8 m/s^2), and h0 is the initial height.
The elastic potential energy stored in the spring at the maximum compression is given by:
Elastic Potential Energy = (1/2) * k * (Δx)^2
where k is the spring constant (320 N/m) and Δx is the maximum compression of the spring.
Since the potential energy at height h0 is equal to the elastic potential energy stored in the spring, we can equate the two expressions:
m * g * h0 = (1/2) * k * (Δx)^2
Substituting the given values:
0.54 kg * 9.8 m/s^2 * h0 = (1/2) * 320 N/m * (0.198 m)^2
Solving for h0:
h0 = ((1/2) * 320 N/m * (0.198 m)^2) / (0.54 kg * 9.8 m/s^2)
= 0.39 m
Therefore, the value of h0 is 0.39 m.
(d) If the block were released from a height 3h0 above the spring, we need to find the maximum compression of the spring.
Using the conservation of mechanical energy principle, we can set the potential energy at height 3h0 equal to the elastic potential energy stored in the spring at the maximum compression:
m * g * (3h0) = (1/2) * k * (Δx_max)^2
Substituting the known values:
0.54 kg * 9.8 m/s^2 * (3 * 0.39 m) = (1/2) * 320 N/m * (Δx_max)^2
Solving for Δx_max:
(1/2) * 320 N/m * (Δx_max)^2 = 5.4 J
(Δx_max)^2 = (5.4 J * 2) / (320 N/m)
= 0.0675 m^2
Δx_max = √0.0675 m^2
≈ 0.26 m
Therefore, the maximum compression of the spring would be approximately 0.26 m.