A 2 000-kg car is slowed down uniformly from 20.0 m/s to 5.00 m/s in 4.00 s. (a) What average force acted on the car during that
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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:
- 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:
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:
- Replacing k and Δx by their values, we get:
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:
- Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:
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