Mark's Struggle to Solve Dynamic Systems Qn

[series111]

[SOLVED] dynamic systems

I Have Just Received This Question = A Car Of Mass 1.2 Tonnes Rolls Down A Hill Which Is 600m Long And Is Inclined At An Angle Of 9 Degrees To The Horizontal. Ignoring The Effects Of Air Resistance And Friction And Assuming The Car To Start From Rest At The Top Of The Hill Calculate Its Speed As It Reaches The Bottom Of The Hill. I Think I Need To Calculate This Using A Vector But I Cant See How I Get The Speed. Please Can Some One Explain This As I Cannot Get My Head Around This Hope Some One Can Help Cheers Mark.

 

[dotman]

Hello,

Have you drawn a free-body diagram? What forces are acting on the car? What equations seem relevant here?

 

[series111]

I Have Done Vectors But I Have No Examples With Another Object This Is What Is Puzzling Me And The Equation Which I Think Is Relevant V=s/t Please Help As I Dont No Where To Start Some Examples Would Be Great Thanks For Your Reply Mark.

 

[dotman]

Hello,

V = s/t is relevant, but I don't think it will help you here. You haven't answered my first two questions.

Have you drawn a free-body diagram, with the forces labeled? If not, do so.

 

[series111]

I Have Drawn A Free Body Diagram With The Forces Labeled And I Am Just Wondering If This Calculation Is Correct 1200 X Sin (9) = 187.72 M/s Or Am I Way Off Thanks For Your Help Again Mark.

 

[cryptoguy]

You can use conservation of energy. PE = KE, so mgh = .5mv^2; h (height) would be 600sin(9).

 

[series111]

Thanks for the advice much appreciated still trying to find my feet in mechnical priciples.

 

[series111]

Is this the correct solution at the top of the system pe = 1.2 tonnes x 9.81 x 93.86 (height) = 1.104 x 10 6 then (pe1+ke1) = (pe2+ke2)
1.104 x 10 6 + 0 = 0 + ke2 ke2 = 1.104 x 10 6 joules
ke = 0.5 mv2

1.104 x 10 6 = 0.5 x 1.2 tonnes x v 2
1.104 x 10 6 = 0.600 v 2
1.104 x 10 6 / 600 = v 2 = 1840
= square root of 1840 = v = 42.89 m/s

 

[cryptoguy]

yep looks right to me. You actually don't even need the mass, it cancels out in PE=KE

 

[series111]

Thanks very much cheers mark

 

[series111]

please can i bother you again as i thought i would be able to answer the following questions after i got the speed here are the rest of the questions (b) calculate its acceleration down the hill (c) the time taken to reach the bottom of the hill (d) the change in the momentum of the car between the top of the hill and the bottom of the hill (e) the change in kinetic energy for the car between the top and bottom of the hill.
also if the brakes are applied at the bottom of the hill slowing the car at a rate of 7.5 m/s2 calculate the stopping distance andthe speed after 5 seconds of braking i am seriously lost as i don't know which equations to use as i have been looking in my notes and several textbooks and still haven't got anywhere your help is much appreciated.

 

[cryptoguy]

b) So you need an equation with Accel. (which you're trying to find), Distance (which you know), Vi and Vf (which you also know). What equation could you use?

c) Simple equation with Vi, Vf, a, and t.

d) Momentum is m*v, so what is it on the bottom compared to what it is on top (hint: on top, v = 0)

e) What is the KE on top? (hint: v = 0 again). And you already calculated the KE on the bottom of the ramp.

For the brakes, you can forget about the hill and just treat this as a 1D motion problem. You know the initial speed and you know the acceleration. You need an equation with Vf, Vi, a, and t again (which you used in part c) and don't forget that since you're braking, the acceleration is negative.

 

[series111]

what does vi and vf stand for as i have the following four formulas

v= u +at u= initial velocity (m/s)

v2 = u2 + 2as v = final velocity (m/s)

s = 1/2 (u +v)t a = acceleration (m/s2)

s = ut + 1/2 at2 s = distance (m)


as for the momentum i calculated 1200 x 42.89 = 51.468 kgm/s but this does not give the change in momentum thanks for your help and quick reply as i am struggling as you can guess but will get there hopefully mark.

 

[cryptoguy]

Sorry about the terminology, Vi is initial velocity, and Vf is final velocity. The momentum on top of the hill is 0 (1200 * 0, right?) so the change in momentum would be the momentum you found - 0 :D

The formulas you have are all you need to solve all parts.

 

[series111]

for the momentum i calculated this the other way last night and i didnt realize it was the same answer mom = m2v - m1u (kgm/s)
(1200x42.89) - (1200x0) = 51.468 kgm/s is this correct ?

as for the acceleration is the formula this a = v2-u2 / 2 x s

also i have just used this formula to work out the kinetic energy ke = 1/2 m (u2-v2)

= 1/2 x 1200 (0 2-42.89 2)
= 1103.73 joules

 

[cryptoguy]

correct

 

[series111]

just to let you know I've been working on these questions today and came up with the following : acceleration a = v2 - u2/2xs = 42.89 2 - 0 2/ 2 x 600 = 1.839 x 10 3 m/s2

time taken t = v - u/a = 42.89 - 0 / 1.839 x 10 3 = 0.023 sec

stopping distance v = u + at = 0 = 42.89 + -7.5

t= v-u/a = 0 - 42.89/-7.5 = 5.71 sec
v = u +at v = 42.89 + -7.5 = 35.39
v = 35.39 x 5.71 = 202.07m

speed after 5 sec braking v = u +at v = 42.89 + -7.5 x 5.71 = 0.065 m/s

thanks again for taking the time to help me yours mark

 

[cryptoguy]

Wait, for the 5 secs braking, why did you use 5.71 for time?

 

[series111]

oh no I've just realized it should be the 5 seconds as that is what it is asking

 

[series111]

just to let you no i used the wrong formula for the stopping distance i should have used a formula with (s) in it not (t) so i have recalculated it using this formula as follows :

v2 = u2+2as

transposed gives you this s=v2-u2/a

so s = 0 squared - 42.89 squared/ - 7.5 = 245.27 m

is this right in your eyes ? thanks for your help mark.