Alrite in my g.11 3u course Im kind of lost. Here are some word problems that I have the answer key to but I have no idea how to get the answer... if anyone can help id be very grateful

1, a shell of mass 1.00 kg is discharged with a speed of 4.5 * 10[exp]2 m/s from a gun having a barrel of length 2.00m

a, how long is the shell in the barrel of the gun
b, calculate the acceleration of the shell while it is in the barrel
c, find the force exerted on the shell while it is in the barrel

2, A horizontal force of 15N pulls a 15.0 kg block along a horizonatal surface. If the force produces an acceleration of 2.0 m/s/s what is the frictioal force acting on the block

3, A 200g rock is suspended by a string. If the tension in the string is 3.2 N find the acceleration of the rock

Okay, I have the first one:

mass:1.00kg
velocity initial: 0m/s
Velocity final:450m/s
distance: 2.00m

d=0.5(vi+vf)t
2=0.5(0+450)t
t=0.008
round to signigicant digits..
The shell was in the barrel for 0.01s

b, vf[exp]2=vi[exp]2 +2ad
450[exp]2=0[exp]2 + 2a(s)
50625=a
round to sig. digits
The acceleration of the shell is 5.5*10[4] m/s/s

c, F=ma
f=1kg(5.5*10[exp]4)
f= 5.5*10[exp]4
The force exerted on the shell while it is in the barrel is 5.5*10[exp]4 Neutons

2 - This is a sum of forces problem. If it's on a horizontal surface, then vertically it's not moving. The vertical forces are the gravitational force (mg) and the normal force (mg), which are equal and opposite. Then, we look at horizontal forces. We know the sum of forces = mass * acceleration. We know there's going to be a frictional force, too. So, we have

15 N - Friction = 15 kg * 2 m/s/s,

and then we should be able to solve for friction


There's a problem with this problem, though. Suppose that we didn't have friction, and we looked at the maximum acceleration we can get from a 15 N force acting on a 15 kg object. We have F=ma, so 15N=15kg * a, so the maximum possible acceleration even without friction is 1 m/s/s. There's no way you could get 2 m/s/s from that force even without friction, much less with it.


3 - Again, this is a question of the sum of forces = ma. The forces acting on the rock are tension and gravity. We take the gravitational force to be positive, so we can write the sum of forces:

(0.2 kg * 9.8 m/s/s) - 3.2 N = (0.2 kg) * a

which gives us a = -6.2 m/s/s. Since we said gravity was positive, this tells us that the rock is accelerating in the direction opposite gravity, or up.