Thread: Kinematics

Can some one help me with 5 ii) a)?

Here is what i've done.

Trailer:
fx = P-250gsin3-150=250a
fy = R-250gcos3=0

Car:
fx = 980-900gsin3-600-P=900a
fy = R-900gsin3=0

so therefore

980-900gsin3-600=P+900a
-250gsin3-150=250a-P

right? simultaneous equation then solve for P and a.
which comes out as:

a=-0.313 ms^-2 (3s.f.)
P=200N

but the answer from the mark scheme says P= 200N but a = 0.713ms^-2 which suggests the equation should be

980+900gsin3-600=P+900a
+250gsin3-150=250a-P

can someone explain what i've done wrong and which one is right.

Sorry for being quite long. THanks for your help.

For the acceleration i got 0.713 m/s^2 also.

((980-750) + (mgsin(3)) = 819.826 then divide by 1150

Originally Posted by Paul46

For the acceleration i got 0.713 m/s^2 also.

((980-750) + (mgsin(3)) = 819.826 then divide by 1150

why is it plus? mgsin(3)

since the weight is going down?

potential energy.

Originally Posted by Paul46

potential energy.

can you explain that?

You have the force plus the PE of which will be converted into KE, so the resultant acceleration will be 0.713m/s^2

Originally Posted by Paul46

You have the force plus the PE of which will be converted into KE, so the resultant acceleration will be 0.713m/s^2

how do i know when to use +/- ?
how come this never came up from previous examples?

Sorry? i'm a bit confused to what you are asking?

Here is a diagram

Originally Posted by Paul46

You have the force plus the PE of which will be converted into KE, so the resultant acceleration will be 0.713m/s^2

how can you turn Gravitational PE into KE in the opposite direction?

question you gave on 5 ii) a) said its going downhill.

Way to go.! lol
thanks!