# Thread: Kinetic and Potential Energy

1. ## Kinetic and Potential Energy

Is there a chance this problem can be solved?

Our Physics Dr. drew this diagram on the board and he said find vo without any further explanation! He stated that it is a complicated problem.

I assume that the object will reach point B with some speed.

I have

Ea = mgh + (1/2)mvo^2

Eb = mgH + (1/2)mv^2 - 100 J

I have many unknowns. I am wondering what is the benefit of the angle! And it is not clear how to find H, it is the height from the ground to point B.

Would it make more sense if I assume there is no speed at point B?

2. ## Re: Kinetic and Potential Energy

Is that a 37 degree launch angle in the upper right corner at what I assume is point B, and does the mass in fact become a projectile at point B until it gets back to point A?

3. ## Re: Kinetic and Potential Energy

Yep 37 degrees. I guess you are Correct. So, how to start?

4. ## Re: Kinetic and Potential Energy

I'm thinking of starting with the projectile motion equations. $\Delta x = 4$ and $\Delta y = 2-H$ ... $\theta = 37^\circ$ indicates a 3-4-5 right triangle, so the values of cosine and sine of $\theta$ will be simple fractions.

I'll have to play with it for a while ...

5. ## Re: Kinetic and Potential Energy

These are the velocity components of point B

Vx = V cos 37°
Vy = V sin 37°

x = xo + V cos 37°

V = 5 m/s

H = 5?

Eb = mgH + (1/2)mv^2 - 100 J= (2 kg)(9.8 m/s^2)(5 m) + (1/2)(2 kg)(5 m/s)^2 - 100 J = 23 J

Ea = mgh + (1/2)mv0^2 = 123 J

v0 = 9.15 m/s

6. ## Re: Kinetic and Potential Energy

Is it correct?

7. ## Re: Kinetic and Potential Energy

x = xo + V cos 37°
How is this equation true without time?

8. ## Re: Kinetic and Potential Energy

t = x/Vcos 37
y = yo + Vsin 37 (t) + (1/2)gt^2
0 = (5) + Vsin 37 (x/Vcos 37) - 4.9 (x/Vcos 37)^2
-5 - xtan 37 = -4.9x^2/V^2 cos^2(37)
V^2(5 + xtan 37) = 4.9x^2/cos^2(37)
V = sqrt[ 4.9x^2/(cos^2(37))(5 + xtan 37) ]
V = sqrt[ 4.9(16)/(cos^2(37))(5 + 4tan 37) ]
V = 3.7 m/s

then

Eb = mgH + (1/2)mV^2 - 100 J = 11.69 J

Ea = mgh + (1/2)mvo^2 = 111.69 J
vo = 8.5 m/s

Tell me. Is it correct NOW?

9. ## Re: Kinetic and Potential Energy

y = yo + Vsin 37 (t) + (1/2)gt^2
0 = (5) + Vsin 37 (x/Vcos 37) - 4.9 (x/Vcos 37)^2
you're saying B is 5 m above ground level? How did you come by that value?

are you also assuming that yf = 0?

using left as positive and point B at (0,5) as you assume, the projectile mass has to pass thru the point (4,2) at a certain time, t, correct?

using your calculated velocity of v = 3.7 m/s ...

$t = \dfrac{\Delta x}{v \cos{\theta}} = \dfrac{4}{3.7\cos(37)} \approx 1.67 \, sec$

$\Delta y = v\sin{\theta} \cdot t - \dfrac{1}{2}gt^2 = 3.7\sin(37) \cdot 1.67 - 4.9 (1.67)^2 \approx -8.7 \, m$

that value of $\Delta y$ doesn't compute ...

10. ## Re: Kinetic and Potential Energy

i made a triangle from point A to point B assuming the angle at A is 37.
if that was not correct. then how can I find the height from point A to point B?

yf should be 2 if initial was 5
that is another mistake

11. ## Re: Kinetic and Potential Energy

Vo = 11.9 m/s

How did you know that the object needs a speed more than 10 m/s? We don't know the height from A to B?

Is it 3?

12. ## Re: Kinetic and Potential Energy

Originally Posted by joshuaa
Vo = 11.9 m/s

How did you know that the object needs a speed more than 10 m/s? We don't know the height from A to B?

Is it 3?
if the mass started at $v_a = 10 \, m/s$ and lost 100 J of energy on the way, at what height would it get to on its way to B?

13. ## Re: Kinetic and Potential Energy

I don't get it if we use 10 m/s cuz we don't know this speed in the beginning
Can I say this

Wf = Fd = 100
d = 100/(2)(9.8) = 5.1 m

distance between A and B = 5.1 - 2 = 3.1 m

14. ## Re: Kinetic and Potential Energy

You keep on guessing ... I'm done.

15. ## Re: Kinetic and Potential Energy

Thanks for the time.