Question

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. a) Find the force constant of the spring. b) Find the total energy of the oscillating system. c) Where will the object be from the equilibrium position when its velocity is -2.0 m/s (negative 2 m/s)

Answer 1

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:

       $F = - k * \Delta x (1)$

• 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:

       $k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

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:

        $U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

• Replacing k and Δx by their values, we get:

       $U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

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:

        $\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

• Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒      $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒     $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$