Question

2. A block of mass 1.2 kg lies on a frictionless surface. A man slides the block

against a spring, compressing it .15m. When the man lets go of the spring, the
block moves at 5 m/s. What is the spring constant of the spring?
mm
.

Answer 1

The spring constant will be "1333.33 N/m".

Step-by-step explanation:

The given values are:

Mass,

m = 1.2 kg

Displacement compression,

x = 0.15 m

Block's velocity,

`v_i=5 \ m/s`

As we know,

⇒
`E_i=E_f`

or,

⇒
`K_i+v_i=K_f+v_f`

⇒
`(1)/(2)mv_i^2+0=0+ (1)/(2)Kx^2`

So,

⇒
`K=(mv_i^2)/(x_2)`

On substituting the values, we get

⇒
`=(1.2* 5* 5)/(0.15* 0.15)`

⇒
`=(30)/(0.0225)`

⇒
`=1333.33 \ N/m`