+1313 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 ΔUg in the gravitational potential energy of the marble-Earth system during the 15 m ascent? (b) What is the change ΔUs in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?
+2353 Okay, so you can find the gravitational potential energy using ΔUg = 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 ΔUg = -ΔUs
(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
+1313 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.
+2353 Okay, so you can find the gravitational potential energy using ΔUg = 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 ΔUg = -ΔUs
(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
