A 7.77- g bullet is moving at 669.00 m/s as it leaves the 0.60- m-long barrel of

Question

A 7.77- g bullet is moving at 669.00 m/s as it leaves the 0.60- m-long barrel of a rifle. What is the average force on the bullet as it moves down the barrel? Assume that the acceleration is constant.

The answer comes out to be a F of 2897.95 N. When I did the calculations my answer was 2897957.5 N. After using v^2=vo^2 + 2a(x-xo) I got an acceleration of 372967.5 m/s^2. I then used F=ma to get the N I just mentioned. Why am having to divide that by 1000 to get the right answer? All of the given numbers are in either meters or seconds. Please help and show work!

Explanation / Answer

v^2 = v0^2 + 2 a x

a = v^2/2x

F = ma = m v^2/(2x) = 7.77E-3*669^2/(2*0.6)=2897.95 N

you forgot to convert to kilograms

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