Question

A 1.60 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 200 N/m and a 285 g metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 14.8 cm below its equilibrium point (call this point A) and released from rest.

Answer 1

The height is 24.03 cm.

Step-by-step explanation:

Given that,

Mass = 1.60 kg

Force constant = 200 N/m

Mass of metal ball = 285 g

Distance x= 14.8 cm

Suppose How high above point A will the tray be when the metal ball leaves the tray?

The force exerted by the spring in the downward

`F=kd`...(I)

The downward force is

`F = mg`...(II)

From equation (I) and (II)

`mg=kd`

`d=(mg)/(k)`

`d=((m_(t)+m_(b))g)/(k)`

Put the value into the formula

`d=((1.60+0.285)*9.8)/(200)`

`d=9.23\ cm`

The point here is 9.23 cm above the equilibrium point and therefore the high above point A is

`d'=d+x`

`d'=9.23+14.8`

`d'=24.03\ cm`

Hence, The height is 24.03 cm.