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

Answer:

The height is 24.03 cm.

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)

$mg=kd$

$d=\dfrac{mg}{k}$

$d=\dfrac{(m_{t}+m_{b})g}{k}$

Put the value into the formula

$d=\dfrac{(1.60+0.285)\times9.8}{200}$

$d=9.23\ cm$

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.