A 6.6 g marble is fired vertically upward using a spring gun. The spring must be compressed 8.7 cm if the marble is to just reach a target 15 m above the marble's position on the compressed spring. **(a)** What is the change Δ*U _{g}* in the gravitational potential energy of the marble-Earth system during the 15 m ascent?

Stu May 15, 2014

#2**+5 **

Okay, so you can find the gravitational potential energy using Δ*Ug = *mgh

where m is mass, g is gravitational force (usually threated as the constant 9.8N/kg ) and h is height

The marble gains all it's potential energy from the spring, therefore Δ*Ug = -Δ Us*

(Technically the spring transfers it's potential energy into kinetic energy after which it becomes potential energy)

The formula for elastic potential energy is given by

$$Pe = -\Delta Us = \frac{1}{2}kx^2$$

where k is the spring constant and x is the amount of compression

edit: I think you can do the last one by yourself if you've finished these

reinout-g May 15, 2014

#1**0 **

Just some help, so I can deduce the answer. I've no real clue tbh.

Also.. A block of mass *m* = 4.0 kg is dropped from height *h* = 32 cm onto a spring of spring constant *k* = 2460 N/m (see the figure). Find the maximum distance the spring is compressed.

Stu May 15, 2014

#2**+5 **

Best Answer

Okay, so you can find the gravitational potential energy using Δ*Ug = *mgh

where m is mass, g is gravitational force (usually threated as the constant 9.8N/kg ) and h is height

The marble gains all it's potential energy from the spring, therefore Δ*Ug = -Δ Us*

(Technically the spring transfers it's potential energy into kinetic energy after which it becomes potential energy)

The formula for elastic potential energy is given by

$$Pe = -\Delta Us = \frac{1}{2}kx^2$$

where k is the spring constant and x is the amount of compression

edit: I think you can do the last one by yourself if you've finished these

reinout-g May 15, 2014