A spring with spring constant k = 810 N/m is placed in a vertical orientation with its lower end supported by a horizontal surface. The upper end is depressed 25 cm, and a block with a weight of 29 N is placed (unattached) on the depressed spring. The system is then released from rest. Assume the gravitational potential energy Ug of the block is zero at the release point (y = 0) and calculate the kinetic energy K of the block for y equal to (a) 0, (b) 0.050 m, (c) 0.10 m, (d) 0.15 m, and (e) 0.20 m. Also, (f) how far above its point of release does the block rise?
To find the kinetic energy (K) of the block at different points during its motion, we need to consider the conservation of mechanical energy.
The system consists of the spring and the block. At the release point (y = 0), the total mechanical energy is the sum of the potential energy due to the spring (Us) and the gravitational potential energy (Ug). At this point, the block is not moving, so its kinetic energy is zero.
The potential energy stored in the spring can be calculated using the formula: Us = (1/2)kx^2
Here, k is the spring constant and x is the displacement of the spring from its equilibrium position.
Given: k = 810 N/m and x = 25 cm = 0.25 m
Us = (1/2)(810 N/m)(0.25 m)^2
Us = 25.3125 J
Since Ug is zero at the release point, the total mechanical energy (E) at this point is given by:
E = Us + Ug
E = 25.3125 J
Now, to find the kinetic energy at different points, we need to subtract the potential energy from the total mechanical energy.
(a) At y = 0:
K = E - Us
K = 25.3125 J - 0 J
K = 25.3125 J
(b) At y = 0.050 m:
To calculate the kinetic energy at this point, we need to find the potential energy when the spring is depressed by 0.050 m.
Us = (1/2)(810 N/m)(0.050 m)^2
Us = 0.2025 J
K = E - Us
K = 25.3125 J - 0.2025 J
K ≈ 25.110 J
(c) At y = 0.10 m:
Us = (1/2)(810 N/m)(0.10 m)^2
Us = 0.810 J
K = E - Us
K = 25.3125 J - 0.810 J
K ≈ 24.502 J
(d) At y = 0.15 m:
Us = (1/2)(810 N/m)(0.15 m)^2
Us = 1.8225 J
K = E - Us
K = 25.3125 J - 1.8225 J
K ≈ 23.490 J
(e) At y = 0.20 m:
Us = (1/2)(810 N/m)(0.20 m)^2
Us = 3.24 J
K = E - Us
K = 25.3125 J - 3.24 J
K ≈ 22.073 J
(f) To calculate how far above its point of release the block rises, we need to find the maximum potential energy it reaches, which occurs at the highest point of its motion. At this point, all the initial potential energy is converted to gravitational potential energy.
Since Ug is zero at the release point and is at a maximum at the highest point, we can find the displacement (y) using:
Ug = mgh
Where m is the mass of the block and g is the acceleration due to gravity.
Given: m = 29 N (weight of the block) and g ≈ 9.8 m/s^2
Ug = mgh
Ug = (29 N)(9.8 m/s^2)h
25.3125 J = (29 N)(9.8 m/s^2)h
Solving for h:
h ≈ 0.091 m
Therefore, the block rises approximately 0.091 m above its point of release.