- #1

Punchlinegirl

- 224

- 0

I used (1/2)mv^2 + mg(S1 sin theta)= FS2

(1/2)(M)(1.93^2) + M (9.8)(.694)= 2.1F

1.86M + 6.81M = 2.1 MuK M(9.8)

Canceling the M's and solving for mk gave me .421, which isn't right.

2.The potential energy stored in the compressed spring of a dart gun, with a spring constant of 28.00 N/m, is .880 J. Find by how much the spring is compressed. I got this part= .251 m. A .050 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position. I got this part= 1.794 m. The same dart is now fired horizontally from a height of 1.70 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.

I used mgh + (1/2)kx^2= (1/2)mv^2

(.050)(9.8)(1.70) + (1/2)(28)(.251^2)= (1/2)(.050)v^2

which gave me 8.28 m/s, which isn't right

Thanks in advance for any help.