Answer:
a) k = 200 N/m
b) E = 4 J
c) Δx = 6.3 cm
Explanation:
a)
- In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:
[tex]F = - k * \Delta x (1)[/tex]
- where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
- Solving for k:
[tex]k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)[/tex]
b)
- Assuming no friction present, total mechanical energy mus keep constant.
- When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:
[tex]U = \frac{1}{2}* k* (\Delta x)^{2} (3)[/tex]
- Replacing k and Δx by their values, we get:
[tex]U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)[/tex]
c)
- When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
- We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
- So, we can write the following expression:
[tex]\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)[/tex]
- Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:
[tex]\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)[/tex]
⇒ [tex]\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)[/tex]
⇒ [tex]\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)[/tex]
