What is the magnitue of the average Force from the wall stopping it?

[nemzy]

What am I doing wrong?

Question:

A 41 g bullet, with horizontal velocity of 496 m/s, stops 13cm within a soid wall. What is the magnitue of the average Force from the wall stopping it?

This is how i tried to solve it

Vi = 496, Vf=0, and Xf-Xi=.13m

So Vf^2=Vi^2+2a(Xf-Xi)

Which = 0 = 246016+2a(.13)

solve for acceleration you get -946215...

F=m(a)...so i did (-946215)(41) which does not equal the answer? Am i doing anything wrong?


2nd Question:

A girl whose weight is 264 N slides down a 5.8 playground slide that makes an angle of 20degree with the horizontal. The coefficient of kinetic friction is 0.10. If she starts at the top with a speed of 0.457 m/s, what is her speed at the bottom?

This is how i tried to slove it

W= change in U + change in K + change in Thermal

Thermal = 143.98 (i won't show my calculations cause i am 100% sure this is righ)
so..
F(d)= -mgh +(1/2)mv^2 - (1.2)m(.457^2) +143.89

After you plug in all your variables, you have two unkowns,velocity and W.. Provided i did the setup right, how do i figure out the velocity of i don't know the magnitude of her force?

 

[jamesrc]

#1: Check your units (41g = .041 kg)

#2. Work-energy theorem:

$$\Delta K + \Delta U = W_{nc}$$
$$K_i = \frac{mv_i^2}{2}$$
$$K_f = \frac{mv_f^2}{2}$$
$$U_f = 0$$ (your choice of datum)
$$U_i = mg(d\sin \theta)$$
$$W_{nc} = -fd$$

where
d = 5.8 m; length of the slide
$$\theta = 20^\circ$$; angle of slide
f = μ N; friction force
μ = 0.10; coefficient of friction
$$N = mg\cos\theta$$; normal force

 

[nemzy]

inst the worky - energy theorem change in K + change in U + change in Thermal (since there is friction?) = W

 

[jamesrc]

OK,

How are you computing the "change in thermal energy" and what is your value for W?

 

[turin]

Originally posted by nemzy
Question:

A 41 g bullet, ...
...
This is how i tried to solve it

... Vf^2=Vi^2+2a(Xf-Xi)
...
solve for acceleration you get -946215...
...

How do you know that the acceleration is constant?

Originally posted by nemzy
2nd Question:

A girl whose weight is ...

I wouldn't do it that way. I would use the kinematic equation that you are trying to use for the first question.