Harvey's Question Thread

[luken93]

F = ma
a = kt/0.5 = 2kt

x = t/2(u +v)
0.8 = t/2(8 + 0)
4/5 = 4t
t = 1/5 sec

x = ut + 1/2at^2
4/5 = 8/5 + 1/2 * 2k/5 * 1/25
4/5 = 8/5 + 2k/250
200 = 400 + 2k
k = -100

Hopefully I haven't made any mistakes

Woops, not constant acceleration :S

[b^3]

I remember doing this question, where is it from (specifically) so I can go and have a look at my notes.

[HarveyD]

my teacher just gave it to us lol
this was what i did

Resultant = 0.5g - kt
m = 0.5 kg
so accel = g - 2kt
vel = gt - kt^2 + 8
v = 0 when x = 80

Setting up simultaneous equations:
0 = gt - kt^2 + 8
80 = g/2kt^2 - 1/3kt^3 + 8t

therefore t = 5.55
and k = 2.02

is that right? :S

[b^3]

I'm sure I have done the exact same question somewhere.
anyway, Yes except it would be 0.8 m for thw displacement since the velocity is in m/s not cm/s. i.e.
k=0.603 and t=17.023
or k=2.585 and t=-0.691
reject the bottom one since t>0

[HarveyD]

hmm maybe he got it off an exam or something

ah yes, thanks!

[HarveyD]

In the Essentials Textbook, they use the term "projected" up an incline, but in the solutions they just use the force acting down the incline i.e. mgsin(theta) to work out the acceleration.

Does that mean we take "projected" as no force being applied....

[luken93]

In the Essentials Textbook, they use the term "projected" up an incline, but in the solutions they just use the force acting down the incline i.e. mgsin(theta) to work out the acceleration.

Does that mean we take "projected" as no force being applied....

Yep, if a force is applied initially, it is never in the force diagram. Crops up every so often in exams as well to trick people.

[b^3]

Projected just means that it is travelling in that direction (although it would have needed to have had a force applied initally to do this), i.e. it has velocity in that direction, so yes projected would mean no force is applied.
EDIT: beaten by 48 seconds.

[HarveyD]

For 25c) of the Chapter 13 Review - Essentials Textbook (Attached)
Is the solution at the back correct, cause i cant seem to get it ><
i keep getting 5/4 rather than 5/4 x g

using this:
u = Answer for b
v = 0
a = -0.2 x g
s = ?

v^2 = u^2 + 2as
so the g's would cancel?

[b^3]

Yeh you're right, It should be 5/4 m not 5g/4 m. The g's cancel.

[HarveyD]

An 800 kg car is subjected to a braking resistance of 60 v Newtons where v = 40 ( 1 - 2e^-0.2t) m/s
Find when it comes to rest:

Teacher did this:
0 = 40 (1 - 2e^-0.2t)

but shouldnt it be

800a = 2400(1-2e^-0.2t)
then anti diff
then solve for when v = 0?

[xZero]

im assuming v is velocity, rest means that velocity is 0 so you let 0 = 40 (1 - 2e^-0.2t), you can do it your way but its gonna take longer