Question

A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is

0.003 kg and the spring constant is 220 N/m. What is the maximum height the marble can reach (a) when

compressed 2 cm? (b) when compressed 4 cm?​

Answer 1

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

$\frac{1}{2}kx^2 = mg(H + x)$

$0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)$

so we have

$H = 1.47 m$

Part b)

Similarly when spring is compressed by 4 cm

then we have

$\frac{1}{2}kx^2 = mg(H + x)$

$0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)$

so we have

$H = 5.94 m$