A 0.51-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 6.7 to 3.3 m/s. What is the spring constant of the spring

487.23 N/m

Explanation:

Given:

mass of metal sphere 'm'= 0.51kg

the spring stretches from 0.12 m to 0.23 m. Therefore,

= 0.12m and = 0.23m

the speed of the sphere decreases from 6.7 to 3.3 m/s. Therefore,

= 6.7m/s and =3.3m/s

In order to find spring constant, we apply law of conservation of energy. i.e

The change of the kinetic energy of sphere is equal to the change of potential energy of the spring.

So, Δ Δ

where,

Δ 1/2 m (- )²

Δ= 1/2 k ( - )²

1/2 m (- )² = 1/2 k (- )²

k= m [ (- )²/(- )²

k= 0.51 [(6.7-3.3)²/ (0.23-0.12)²]

k= 487.23 N/m

Thus, the spring constant of the spring is 487.23 N/m

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2