Question

A 5.0-kg box moving at 6.0 m/s on a horizontal, frictionless surface runs into a light spring of force constant 60 N/cm. You may want to review (Pages 183 - 189) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Motion with a varying force.

Answer 1

Maximum compression distance of spring = 0.5 cm

Step-by-step explanation:

Hooke's law of spring expansion states that the force exerted on the spring is proportional to the spring constant. In mathematical terms, this can be expressed as:

`F = kE`

where F = Force on the spring

k = spring constant

E = length of stretch or compression.

but force = Ma

F = ma

= 5 * 6.0

= 30 N

Using Hooke's law

F = kE

30 = 60 E

E = 0.5 cm

So the spring is compressed by 0.5 cm

Answer 2

The maximum compression of the spring is 1.73 cm.

Step-by-step explanation:

Given that,

Mass of box = 5.0 kg

Speed = 6.0 m/s

Spring constant = 60 N/cm

Suppose use the work-energy theorem to find the maximum compression of the spring.

We need to calculate the maximum compression of spring

Using work -energy theorem

`(1)/(2)mv^2=(1)/(2)kx^2`

`x^2=(mv^2)/(k)`

`x=\sqrt{(mv^2)/(k)}`

Put the value into the formula

`x=\sqrt{(5.0*(6.0)^2)/(60)}`

`x=1.73\ cm`

Hence, The maximum compression of the spring is 1.73 cm.