a 5kg block on a rough horizontal surface is attached to a light spring (force constant=1.6kN/m). the block passes through its e

Question

3 years ago

10

a 5kg block on a rough horizontal surface is attached to a light spring (force constant=1.6kN/m). the block passes through its e

quilibrium position with a kinetic energy of 5J and is brought momentarily to rest after stretching the spring .06m. how much work is done by the frictional force on the block as it moves from its equilibrium position to the point of momentary rest?

Physics

1 answer:

natima [27] · Answer 1 · 2022-06-28T04:10:45+03:00

natima [27]3 years ago

5 0

Answer:

2.12 J

Explanation:

Initial kinetic energy = final elastic energy + work by friction

KE = EE + W

KE = ½ kx² + W

5 J = ½ (1600 N/m) (0.06 m)² + W

W = 2.12 J