kinematics question, need help!

[yabbaboo]

A box of mass 68kg and attached to a mass of 28kg through a smooth pulley is held at rest on a smooth sloping plane inclined at 20 degrees to the horizontal. The mass is then let go causing it to slide. Find the direction of movement of the box, the acceleration of the mass and the distance it will move in 3.2 seconds. The diagram is attached.

I got the answer but it is wrong. The answer is 2.46 m. Can anyone help me please?

[b^3]

Draw the forces on the diagram that is acting.
That is
On the 28 kg mass
28g N down
T in the rope up (tension)
On the 68 kg mass
T in the rope from the 68 kg mass
68g N vertically down on the 68 kg mass
and R (reaction force) perpendicular to the surface

Now next you need two simulataneous equations.
So on the 28 kg mass
T-28g=ma
T-28g=28a....1
On the 68 kg mass -  Take down the slope as positive.
Resolving peroendicular to the slope - R-68gcos(20)=0
R=68g*cos(20) - Not needed here since no friction
also
Resolving paralell to the slope - 68g*sin(20)-T=68a....2
From 1
T=28a+28g
sub into 2
68gsin(20)-28a-28g=68a
g(68sin(20)-28)=96a
g=9.8
a=-0.4814 m/s^2

So that is up the slope. (since down the slop was taken as positive and the acceleration turned out to be negative, so it is accelerating in the opposite direction).
now using our equ of motions
a=0.4814 m/s^2
u=0m/s
x=?
t=3.2s
x=ut+1/2*at^2
x=(0*3.2)+1/2 *(0.4814)*(3.2)^2
x=2.479m
x=2.48m

So it travels up the plane and after 3.2 seconds it has travelled 2.48m.

The answer is slightly out due to rounding somewhere, I'll see if I can find it. EDIT 2: Can't seem to fix it, maybe the answer is slightly wring, unless I've made a mistake.

EDIT: added image.

[Bhootnike]

Draw the forces on the diagram that is acting.
That is
On the 28 kg mass
28g N down
T in the rope up (tension)
On the 68 kg mass
T in the rope from the 68 kg mass
68g N vertically down on the 68 kg mass
and R (reaction force) perpendicular to the surface

Now next you need two simulataneous equations.
So on the 28 kg mass
T-28g=ma
T-28g=28a....1
On the 68 kg mass -  Take down the slope as positive.
Resolving peroendicular to the slope - R-68gcos(20)=0
R=68g*cos(20) - Not needed here since no friction
also
Resolving paralell to the slope - 68g*sin(20)-T=68a....2
From 1
T=28a+28g
sub into 2
68gsin(20)-28a-28g=68a
g(68sin(20)-28)=96a
g=9.8
a=-0.4814 m/s^2

So that is up the slope. (since down the slop was taken as positive and the acceleration turned out to be negative, so it is accelerating in the opposite direction).
now using our equ of motions
a=0.4814 m/s^2
u=0m/s
x=?
t=3.2s
x=ut+1/2*at^2
x=(0*3.2)+1/2 *(0.4814)*(3.2)^2
x=2.479m
x=2.48m

So it travels up the plane and after 3.2 seconds it has travelled 2.48m.

The answer is slightly out due to rounding somewhere, I'll see if I can find it. EDIT 2: Can't seem to fix it, maybe the answer is slightly wring, unless I've made a mistake.

EDIT: added image.

I'm probably wrong,  but shouldn't the weight force acting down on the weights be w= mg?  So if g = 10,  ( or 9.,  w= 280. And 680. N. Respectively?

[Lasercookie]

I'm probably wrong,  but shouldn't the weight force acting down on the weights be w= mg?  So if g = 10,  ( or 9.,  w= 280. And 680. N. Respectively?

Draw the forces on the diagram that is acting.
That is
On the 28 kg mass
28g N down
T in the rope up (tension)
On the 68 kg mass
T in the rope from the 68 kg mass
68g N vertically down on the 68 kg mass

28g N = 28 * g N= 28 * 10 N = 280 N

He has considered w=mg, only that he didn't substitute in the value for g, instead leaving it as 28g and 68g.

[b^3]

Draw the forces on the diagram that is acting.
That is
On the 28 kg mass
28g N down
T in the rope up (tension)
On the 68 kg mass
T in the rope from the 68 kg mass
68g N vertically down on the 68 kg mass
and R (reaction force) perpendicular to the surface

Now next you need two simulataneous equations.
So on the 28 kg mass
T-28g=ma
T-28g=28a....1
On the 68 kg mass -  Take down the slope as positive.
Resolving peroendicular to the slope - R-68gcos(20)=0
R=68g*cos(20) - Not needed here since no friction
also
Resolving paralell to the slope - 68g*sin(20)-T=68a....2
From 1
T=28a+28g
sub into 2
68gsin(20)-28a-28g=68a
g(68sin(20)-28)=96a
g=9.8
a=-0.4814 m/s^2

So that is up the slope. (since down the slop was taken as positive and the acceleration turned out to be negative, so it is accelerating in the opposite direction).
now using our equ of motions
a=0.4814 m/s^2
u=0m/s
x=?
t=3.2s
x=ut+1/2*at^2
x=(0*3.2)+1/2 *(0.4814)*(3.2)^2
x=2.479m
x=2.48m

So it travels up the plane and after 3.2 seconds it has travelled 2.48m.

The answer is slightly out due to rounding somewhere, I'll see if I can find it. EDIT 2: Can't seem to fix it, maybe the answer is slightly wring, unless I've made a mistake.

EDIT: added image.

I'm probably wrong,  but shouldn't the weight force acting down on the weights be w= mg?  So if g = 10,  ( or 9.,  w= 280. And 680. N. Respectively?

It is, just instead of using 280 and 680, I keep it as 28g and 68g and subbed g back in at the end.

EDIT: yes as laseredd has said.

[Bhootnike]

Oh ok,  well there you go
Maybe cause you used 9.8 it came out a bit off the answer? 

[b^3]

Oh ok,  well there you go
Maybe cause you used 9.8 it came out a bit off the answer? 

I tried g=10 m/s^2 as well and it was off by more.

[dc302]

You should always use g=9.8 in spesh.

[b^3]

You should always use g=9.8 in spesh.

Yeh make sure you always check the front of the section, where it has the intructions. It will tell you what value of g to use. For spesh g=9.8m/s^2 (as dc302 said) and for physics its g=10 m/s^2. But always check just in case.

[Lasercookie]

If you round off a to 0.48, you get
That's my guess to whatever the source of the question has done. I would think that's excessive rounding off though.

[b^3]

If you round off a to 0.48, you get
That's my guess to whatever the source of the question has done. I would think that's excessive rounding off though.

Good spot laseredd.

[dc302]

Lol I know I'm just being pedantic but g is (at least in spesh) used as the number, not the acceleration. Ie, g=9.8, not 9.8 m/s^2

[b^3]

Lol I know I'm just being pedantic but g is (at least in spesh) used as the number, not the acceleration. Ie, g=9.8, not 9.8 m/s^2

Yes thats right, and being pedantic pays off in the end, small things make the difference between a 50 and a 49.

EDIT: Just note g=9.8 while the acceleration due to gravity = 9.8m/s^2

[Lasercookie]

Lol I know I'm just being pedantic but g is (at least in spesh) used as the number, not the acceleration. Ie, g=9.8, not 9.8 m/s^2

Nothing wrong with being pedantic

But what is the reasoning behind not writing m/s^2?

[b^3]

From VCAA 2010 specialisr exam

Take the acceleration due to gravity to have magnitude g m/s², where g=9.8.

So yeh as said in previous post.