An object of mass M = 4 kg slides from rest a distance d = 2 m down a frictionless inclined plane where it encounters a spring. It compresses the spring a distance D = 1.2 m before stopping. The inclined plane makes an angle q = 20° with the horizontal.
a) When the block just touches the spring, how much gravitational potential energy has it lost? Give your answer as a positive number.
D|Ugrav,1|= J
b) After the mass has fully compressed the spring, what is its total loss of gravitational potential energy? Give your answer as a postive number.
D|Ugrav,2| = J
c) What is the value of the spring constant?
To calculate the answers to these questions, we need to break down the problem into different steps. Let's go through each step:
a) When the block just touches the spring, it has not yet compressed it. Therefore, no gravitational potential energy has been converted or lost. Thus, D|Ugrav,1| = 0 (Joules).
b) After the mass has fully compressed the spring, it has come to a stop. At this point, all of its initial gravitational potential energy has been converted into the potential energy stored in the compressed spring. To calculate the total loss of gravitational potential energy, we need to find the initial and final gravitational potential energy of the object.
The initial gravitational potential energy (Ugrav,i) can be calculated using the formula: Ugrav,i = m * g * h.
Given that the mass (m) is 4 kg and the height (h) can be found using the inclined plane's angle (q) and the distance (d), we have:
h = d * sin(q)
Plugging in the values, we have:
h = 2 m * sin(20°) ≈ 0.68 m
Ugrav,i = 4 kg * 9.8 m/s^2 * 0.68 m ≈ 26.95 J
The final gravitational potential energy (Ugrav,f) is equal to the gravitational potential energy at the point where the object stops, which is when it is fully compressed against the spring.
To calculate Ugrav,f, we need to find the vertical displacement (s) of the object. This can be calculated using the inclined plane's angle (q) and the compression distance of the spring (D) as follows:
s = D * cos(q)
Plugging in the values, we have:
s = 1.2 m * cos(20°) ≈ 1.12 m
Ugrav,f = m * g * s = 4 kg * 9.8 m/s^2 * 1.12 m ≈ 44.15 J
The total loss of gravitational potential energy (D|Ugrav,2|) is the difference between the initial and final gravitational potential energies, so:
D|Ugrav,2| = Ugrav,i - Ugrav,f ≈ 26.95 J - 44.15 J ≈ -17.2 J
Note that the negative sign indicates a loss of potential energy.
c) To find the spring constant (k), we can use Hooke's Law, which states that the force required to compress or extend a spring is linearly proportional to the displacement. Mathematically, it is represented as:
F = -k * x
Where F is the force applied to the spring, x is the displacement, and k is the spring constant.
In this case, the force exerted by the spring when it is fully compressed is equal to the weight of the object, which can be calculated using the formula:
F = m * g
Plugging in the values, we have:
F = 4 kg * 9.8 m/s^2 ≈ 39.2 N
Now we can solve for the spring constant (k):
39.2 N = -k * 1.2 m
Rearranging the equation:
k = -39.2 N / 1.2 m ≈ -32.67 N/m
The negative sign indicates that the force exerted by the spring is in the opposite direction of the compression. The spring constant (k) is approximately 32.67 N/m.