A mass of 0.755 kg is attached to a horizontal spring of constant 403 N/m as shown. The system is compressed a distance xa = 20.8 cm from its equilibrium position and released from rest. The coefficient of friction on the surface is μ = 0.354. The mass comes to rest again at some distance xc on the other side of the equilibrium position.
A. What is the elastic potential energy at point A?
B. What is the deformation of the spring at point C?
A. elastic P.E. = (1/2) k Xa^2
Xa must be in meters, since k has units of N/m.
B. The difference in elastic P.E at the two end positions is the work done against friction.
(1/2)k [Xa^2 - Xc^2) = M g ì (Xa + Xc)
Solve for Xc
To determine the answers to these questions, we'll need to consider the energy of the system at each point.
A. To find the elastic potential energy at point A, we can use the formula for the elastic potential energy stored in a spring:
Elastic potential energy = (1/2) * k * x^2
where k is the spring constant and x is the deformation or displacement from the equilibrium position.
Given:
Mass (m) = 0.755 kg
Spring constant (k) = 403 N/m
Deformation (xa) = 20.8 cm = 0.208 m
Substituting these values into the formula:
Elastic potential energy at point A = (1/2) * 403 * (0.208)^2
Calculating this expression gives the answer for part A.
B. To find the deformation of the spring at point C, we can use the conservation of mechanical energy principle. At the equilibrium position, the total mechanical energy of the system will be equal to the mechanical energy at point A.
The total mechanical energy of the system is given by the sum of the kinetic energy (KE) and potential energy (PE):
Total mechanical energy = KE + PE
At point A, the mass is at maximum deformation, so its velocity is zero, and therefore the KE is zero. Hence,
Total mechanical energy at point A = PE at point A
At point C, the mass comes to rest again, meaning its final kinetic energy is zero. Therefore:
Total mechanical energy at point C = PE at point C
Now, we need to calculate the potential energy at point C:
PE at point C = m * g * h + (1/2) * k * x^2
where m is the mass, g is the acceleration due to gravity, h is the displacement from the equilibrium position at point C, and x is the deformation of the spring.
Since the mass comes to rest again, the potential energy at point C will be equal to the elastic potential energy stored in the spring. Hence,
Elastic potential energy at point C = (1/2) * k * x^2
We can equate the total mechanical energy at point A (which is equal to the potential energy at point A) to the total mechanical energy at point C (which is equal to the elastic potential energy at point C) and solve for x.
Substituting the given values into the equation and solving for x will give us the answer for part B.