# Incline problem

1. Jun 10, 2007

### pharaoh

angle of incline is 45 degree, length of the incline is 30m, and the the time of drop is 2.14

1- what should your velocity be at the base of the drop (frictionless)?

v=30/2.14=14.0*2=28m/s

2- what is you velocity at the base of the drop?
mgh=1/2mv^2
2gh=v^2
2(9.81)(30)=v^2
square root of 588.6=24.26m/s

what should the acceleration be for #1?

ma=mgsin45
a=gsin45
a=9.81*sin45= 6.937m/s^2

what should your acceleration be for #2?
a=v/t=24.26/2.14=11.33/2= 5.75m/s^2

am I right?

Last edited: Jun 10, 2007
2. Jun 10, 2007

### Mindscrape

Er, what exactly is the problem?

3. Jun 10, 2007

### pharaoh

the problem is: from the rest you slide on an incline with an angle of 45 degree and the length of the incline is 30 meter and the time of the drop is 2.14 seconds:

1- what should your velocity be at the base of the drop (frictionless)?

2- what is you velocity at the base of the drop?

what should the acceleration be for #1?

what should your acceleration be for #2?

4. Jun 10, 2007

### G01

You have made a few mistakes but your on the right track.

Your answer for part 1 is wrong. Since v isn't constant, your can't use v=d/t. Try a different kinematic equation.

For part 2: This is set up correctly if there is no friction. So, you can actually use this method for part 1. The only problem is that 30m is not the height of the object, it is the length of the incline. Use trig to find the height. For this part, you have to work out the acceleration with friction(answer to part 4). Use the time to find the actual acceleration with friction involved.

For part 3: This is correct.

Part 4: This will be correct assuming you find the correct final velocity in part 2 and the object starts with velocity of 0.

Good luck! See how far you can get now!

Last edited: Jun 10, 2007
5. Jun 10, 2007

### pharaoh

for number one
d=1/2at^2
d=.5(9.81)(2.14^2)=22.46 meter (vertical drop)

2gh=v^2
2(9.81)(22.46)=V^2
v= 20.99 m/s

i couldn't figure out number 2

6. Jun 10, 2007

### G01

That method for finding the height won't work. Use trigonometry, Make a triangle with the slope. otherwise that's correct. I got to go for now, but if you still haven't got #2 answered I'll help you when I get back. Good Luck!

7. Jun 10, 2007

### pharaoh

since it 45, 45, 90 triangle
30/the square root of 2= 21.213 meter
2(9.81)(21.213)=v^2
v=20.40

what about number two, I couldn,t figure it out

8. Jun 10, 2007

### pharaoh

for number two:
v=30/2.14=14*2=28

Vf=Vi+at=
a= 28/2.14=13.0/2= 6.55 m/s^2
v=6.55*2.14=14 m/s

Last edited: Jun 10, 2007
9. Jun 10, 2007

### G01

This gives you the average speed, which probably isn't exact enough for the problem. Do you know the coefficient of friction for the slope?

10. Jun 10, 2007

### pharaoh

the coefficient of friction= Fn/Fg=Fg sin (theta)/Fg cos(theta)=Tan (theta)
=tan 45= 1

the coefficient is not given so i tried to solve it

11. Jun 10, 2007

### pharaoh

can i use this formula for the acceleration then i can find the velocity

a= gsin(theta)+Mkgcos(theta)
since mk=tan(theta)

12. Jun 10, 2007

### pharaoh

for the acceleration
Vf= V.*t+0.5at^2
Vf=d+0.5at^2
a=30/05(2.14^2)=13.10 m/s^2

13. Jun 10, 2007

### G01

That can't be the coefficient of friction. That equation is only true if the force of gravity equals the force of friction, which we know it doesn't since the object still has a net force accelerating it. Are your sure more information wasn't given?

14. Jun 10, 2007

### pharaoh

for the acceleration
Vf= V.*t+0.5at^2
Vf=d+0.5at^2
a=30/05(2.14^2)=13.10 m/s^2

15. Jun 10, 2007

### G01

That won't work since v is not constant. The only time you can apply v=d/t without error is when v is constant. Try this:

Use $$x_f=x_o+v_ot+.5at^2$$

Here you start at x_o=0 and end at x_f=30. You know the time. Go from here.

16. Jun 10, 2007

### pharaoh

a=30/0.5(2.14^2)
because Vi is zero and the x_o is zero also

17. Jun 10, 2007

### pharaoh

what formula will make me fined the coefficient since the Tan (theta) is wrong

18. Jun 10, 2007

### G01

That looks good for the acceleration. Now using a and t you can easily find v_f for part 2.

Off the top of my head, the only way I can think of finding the coefficient right now is to find the change in net force between the two scenarios. This change will be equal to the friction force. You can then use this value to find the coefficient. It's not part of the problem you posted, but you can try it if you want practice!

Otherwise, good job getting this problem!

Last edited: Jun 10, 2007
19. Jun 10, 2007

### pharaoh

thanks that help, I really appreciate it

20. Jun 10, 2007

### G01

No problem! Good luck in the future!