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Physics question

F

FFactory0x

Diamond Member
2 physics questions. Need formula and how to do each. Having much trouble

A ball is rolling down a incline with a constant acceleration. Took .4s to cover the first .32 m. How long did it take to cover the next .180




A bullet travelling at 120m/s tears into a slab of thick wood. The wood exerts negative accel. of 8000 m/s/s. How long will it take the bullet to stop

Still try to figure out #1

Heres #3

If acceleration produced by a force of 50N on an acting mass of 10kg moving on a table is 3.0^2. What is the magnitude of the frictional force between the mass and the table
 
Originally posted by: FFactory0x
would a=10
Click to expand...

I should be 9.8 meters/second^2

if you knew the mass of the bullet you could use the conservation of energy to solve the prolbem .


or mad= 1/2mv^2 where m is the mass of the bullet, a is the negative accel of the bullent, v is velocity and d is what you are solving for.

 
a=v/t
v=d/t

Find v at .4s, then find a at .4s. You know that a is constant, so your value for a at .4s is the same for the entire trajectory. Then use Heisenberg's equation.
 
You will find A in this case, you won't use 10. You use 10 when you're talking about only gravity. This is an inclined plane, and the acceleration trajectory you're talking about is an angled vector... not free fall.
 
man im still lost. Could someone right out the euation for each along with numbers plugged in so i can see how its done and understand the concept better
 
1)
Edit: Wrong answer

2) v = v0 + a*t
v = 0 since we want to know how long it will take for the bullet to stop
v0 = -a*t
120 m/s = -(-8000 m/s^2) * t
Solve for t to get 0.015s
 
Actually, that's wrong. You forgot to take into account the fact that the ball was already rolling quite fast at .4 seconds. The equation should be:

d = v * t + .5 * a * t * t

Okay, so the actual time is:

.18 = 4 * .4 * t + .5 * 4 * t * t
.18 = 1.6t + 2t*t

t*t + .8t - .09 = 0

<quadratic formula magic>

t = .1 (or -.9, which doesn't make sense)

That's one of your options and the numbers worked out quite nicely, which leads you to believe it's right.
 
Here's an easier way to do it. You've found acceleration at 4 m/s^2. If this is true, then:

T(.32)=.4s given (distance traveled at .4s is .32m)
T(.32+.180) = .5s -- plug in .500=(.5)(4)(t^2), solve for t.

Subtracting your time at .32 meters from your time at .500 meters gives you your time of .1 seconds.
 
Originally posted by: dornick
Actually, that's wrong. You forgot to take into account the fact that the ball was already rolling quite fast at .4 seconds. The equation should be:

d = v * t + .5 * a * t * t

Okay, so the actual time is:

.18 = 4 * .4 * t + .5 * 4 * t * t
.18 = 1.6t + 2t*t

t*t + .8t - .09 = 0

<quadratic formula magic>

t = .1 (or -.9, which doesn't make sense)

That's one of your options and the numbers worked out quite nicely, which leads you to believe it's right.
Click to expand...
Yeah, that looks right. You have to add the velocity it picks up during the first 0.4s.
 
Use F=ma on question 3. That's the only equation you need, plus a subtraction to isolate actual from ideal.
 
Total Force = Force(produced by friction) + Force(of acceleration)

Force of acceleration is 10kg*3m/s^2 = 30.

So the friction component is 20N.
 
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