joshik0506
contestada

A block of mass 0.436 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.634 m.

Find the spring constant.

The block is then pulled down an additional 0.382 m and released from rest. Assuming no damping, what is its period of oscillation?

How high above the point of release does the block reach as it oscillates?

A block of mass 0436 kg is hung from a vertical spring and allowed to reach equilibrium at rest As a result the spring is stretched by 0634 m Find the spring co class=

Respuesta :

MathPhys

Answer:

6.74 N/m

1.60 s

0.764 m

Explanation:

There are two forces on the block, weight mg pulling down and spring force kx pulling up.

mg = kx

(0.436 kg) (9.8 m/s²) = k (0.634 m)

k = 6.74 N/m

The period of a mass on a spring is:

T = 2π √(m / k)

T = 2π √(0.436 kg / 6.74 N/m)

T = 1.60 s

The amplitude of the oscillation is 0.382 m, so the block will reach a height of 2 × 0.382 = 0.764 m above its lowest point.

Otras preguntas