A spring gun is loaded with a 500-gram projectile. The spring is massless and th

Question

A spring gun is loaded with a 500-gram projectile. The spring is massless and therefore has no kinetic energy.

Note: I'm in an algebra-based physics class, so please answer with algebra-based solutions. Please help! I'm pulling out my hair trying to figure this out.

1. How much potential energy is converted to kinetic energy in the spring gun? Use scientific notation in the format X.XXe-X, with three sig figs as indicated. (Note: At the start t=0 s; x=-5.0 cm; v=0 m/s. At the end of the run t=4.0 s; x=10.03 cm; v=4.47 m/s.) Answer in Joules.

2. How much PE has been converted to KE when the ball is at -2.99 cm? (Note: at -2.99 cm, the time is 1.04 sec and the velocity is 3.59 m/s.) Answer in Joules.

3. How much PE has been converted to KE when the ball is at -1.98 cm? (Note: at -1.98 cm, the time is 1.30 sec and the velocity is 4.10 m/s.) Answer in Joules.

4. How much PE has been converted to KE when the ball is at -0.96 cm? (Note: at -0.96 cm, the time is 1.54 sec and the velocity is 4.39 m/s.) Answer in Joules.

5. How much PE has been converted to KE when the ball is at 0 cm? (Note: at 0 cm, the time is 1.76 sec and the velocity is 4.47 m/s.) Answer in Joules.

Explanation / Answer

isn't kinetic energy = 0.5*m*v^2
thats it. Use this formula everywhere with m =0.5 Kg

1.
K.E = 0.5*m*v^2
= 0.5*0.5*4.47^2
=5.00 J

2.
K.E = 0.5*m*v^2
= 0.5*0.5*3.59^2
=3.22 J

3.
K.E = 0.5*m*v^2
= 0.5*0.5*4.10^2
=4.20 J

4.
K.E = 0.5*m*v^2
= 0.5*0.5*4.39^2
=4.82 J

5.
K.E = 0.5*m*v^2
= 0.5*0.5*4.47^2
=5.00 J

I have considered each question as independent and solved it.
If they are related, answer will be different

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