What force, in Newtons, does the wood exert on the bullet?

[Whatupdoc]

A .22 rifle bullet, traveling at 350 m/s, strikes a block of soft wood, which it penetrates to a depth of 0.130 m. The block of wood is clamped in place and doesn't move. The mass of the bullet is 1.80 g. Assume a constant retarding force.

1.) What force, in Newtons, does the wood exert on the bullet?

using Newton's Second Law (F=ma)

first i had to find a
v^2f = V(0)^2 + 2ax <--- equation of motion
0=350^2+2(a)(0.130)
a = 471153.8462

then plug in known values for Newton's second law

F = (1.80)(471153.8462) = 848076 N
or 8.48×10^5 N

but it's the wrong answer, can someone help?

 

[Nylex]

You've not converted your mass to kilograms.

 

[M98Ranger]

The force exerted by the wood on the bullet can be calculated using the equation F=ma, where F is the force, m is the mass of the bullet, and a is the acceleration. In this case, the acceleration can be calculated using the equation v^2f = v0^2 + 2ax, where v0 is the initial velocity (350 m/s) and x is the distance the bullet penetrated (0.130 m). Solving for a, we get a= 471153.8462 m/s^2.

Now, we can plug in this value for a into the equation F=ma. Therefore, the force exerted by the wood on the bullet is F= (1.80 g)(471153.8462 m/s^2) = 848076.923 N or 8.48×10^5 N.

It is important to note that this is the maximum force exerted by the wood on the bullet, as it is assumed that there is a constant retarding force acting on the bullet as it penetrates the wood. This means that the actual force may be slightly less than this calculated value.