Question

A block with mass of 5.00 kg is attached to a horizontal spring with spring constant of 400 N m-1, as in the figure beside. The surface the block rests upon is frictionless. If the block is pulled out to xi = 0.05 m and released,find the speed of the block when it first reaches the equilibrium point

Answer 1

Answer:

3 kg is right fine happy ok

Answer 2

Explanation:

Given A block with mass of 5.00 kg is attached to a horizontal spring with spring constant of 400 N m-1, as in the figure beside. The surface the block rests upon is frictionless. If the block is pulled out to xi = 0.05 m and released, Find the speed of the block when it first reaches the equilibrium point.

• So a block of mass 5 kg is attached to a horizontal spring with spring constant 400 N/m.
• The block is pulled out to xi = 0.05 m and released.
• By conservation of energy we have
•                                 ½ kxi^2 = ½ mv^2 + ½ kxo^2
•                                            here xo = 0
•                                   So ½ kxi^2 = ½ mv^2
•                                    So V = xi √k/m
•                                          V = 0.05 √400 / 5
•                                          V = 0.05 x √80
•                                            V = 8.9442 x 0.05
•                                            V = 0.4472 m/s

Reference link will be

https://brainly.in/question/3894042