Question

2. A 0.5 kg block lies in a frictionless horizontal plane connected to an ideal spring. The block is pull to the right so that the block is displace by 0.2 m from the equilibrium point. The block then released with initial speed 1 m/s. When it reaches the equilibrium point, the block moves with speed 2 m/s. Determine: a. Total mechanical energy of the system! b. The spring constant! c. How far the spring is compressed as the block moves to the left!

          2. A 0.5 kg block lies in a frictionless horizontal plane connected to an ideal spring. The block is pull to the right so that the block is displace by 0.2 m from the equilibrium point. The block then released with initial speed 1 m/s. When it reaches the equilibrium point, the block moves with speed 2 m/s. Determine:
a. Total mechanical energy of the system!
b. The spring constant!
c. How far the spring is compressed as the block moves to the left!

2. A 0.5 kg block lies in a frictionless horizontal plane connected to an ideal spring. The block is pull to the right so that the block is displace by 0.2 m from the equilibrium point. The block then released with initial speed 1 m/s. When it reaches the equilibrium point, the block moves with speed 2 m/s. Determine:
a. Total mechanical energy of the system!
b. The spring constant!
c. How far the spring is compressed as the block moves to the left!

Added by Javad R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Jacob Fry

10/06/2022

Step 1

Given: Mass, m = 0.5 kg Initial speed, v = 1 m/s Final speed, v' = 2 m/s Total mechanical energy = Kinetic energy at equilibrium point Total mechanical energy = 0.5 * 0.5 * (2)^2 Total mechanical energy = 1 Joule Show more…

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