Question

The spring in a toy pop-gun has a spring constant of 85 N/m. To push the cork out of the barrel, the spring is compressed by 0.030 m.

Part 1: Calculate the elastic potential energy of the cork.

Part 2: Find the speed of the 0.002 kg cork as is leaves the barrel of the toy pop-gun.

**Assume no losses from friction. How would I solve and show the work for this? :/

Thank you! :)

Answer 1

hint:
the formula which gives the requested potential energy is:
\[\Large EP = \frac{1}{2}k{\delta ^2}\]
where k= 85 N/m, and \delta=0.03 meters

Answer 2

in order to answer ot part B, we have to keep in mind that all the potential energy will transform into kinetic energy of the cork, namely:
\[\Large EP = KE = \frac{1}{2}m{v^2}\]
where m is the mass of the cork, and v is the requested velocity.
EP is the potential energy computed in part A.. So you have to solve this equation:
\[\Large v = \sqrt {\frac{{2EP}}{m}} = ...?\]

Answer 3

oops.. all the potential energy will transform itself into kinetic energy

Answer 4

so for part 1, i get 1/2(85)(0.03) = 1/2(2.55) = 1.275 ? for the potential energy?

and for part 2, i am confused as to what to plug in? @Michele_Laino

Answer 5

part 1
the right computation is:
\[\Large PE = \frac{1}{2} \times 85 \times {\left( {0.03} \right)^2} = \frac{{85 \times 9}}{{200}} = 3.825\;Joules\]

Answer 6

so, we can write:
\[\Large v = \sqrt {\frac{{2PE}}{m}} = \sqrt {\frac{{2 \times 3.825}}{{0.002}}} = ...?\]

Answer 7

ohh okay! i forgot to square it in part 1! so 3.825 J would be the final solution? for part 1? that is the potential energy of the cork?

and then for part 2, calculated, v=61.84658438? so the final solution for part 2 is the speed (velocity) is 61.85?
not sure about the units though :/ would it be m/s ? :/ @Michele_Laino

Answer 8

that's right!

Answer 9

yay!! thank you!! :)

Answer 10

thank you!! :)