# M1Watch

#1
Very basic, just want to check answer with u guys;

A block of mass 2kg slides on smooth horizontal table. It is attached to hanging object mass 0.5kg by inelastic string that presses over pulley. Block is released;

i) calculate acceleration of block ii) tension in string

thanks
0
12 years ago
#2
(Original post by Dev.420)
Very basic, just want to check answer with u guys;

A block of mass 2kg slides on smooth horizontal table. It is attached to hanging object mass 0.5kg by inelastic string that presses over pulley. Block is released;

i) calculate acceleration of block ii) tension in string

thanks
i)f = ma
0.5g = 2*a
a = 0.5g/2
ii)0.5g-T=0.5*0.5g/2
make T the subject and you should get the answer! Hope that helped!
0
12 years ago
#3
I'm confused, when I read this I said:

both objects have to move with the same acceleration. So, the block: T = 2a, and the ball, 0.5g - T = 0.5a

Adding the two eqs, 0.5g = 2.5a, a = 1.96

So T = 2 * 1.96 = 3.92N.

But we get different answers. Am I wrong? Why does this not work?
0
12 years ago
#4
I think our method is right. I don't know how Coolraj thinks he worked out the tension...

Let's sit tight and wait for someone else to come along and clarify.
0
12 years ago
#5
i get a=1.96 too, and so T=3.92 ...
see, with these kind of questions, i reckon they wouldn't ask for the acceleration unless they wanted an equation that looked like mg-T=ma . cos if it's all in equilibrium, it's quite simple as then T=mg, and there's no a anyway.
so i assume the clue is in the Q =P
0
12 years ago
#6
(Original post by guitargirl9)
see, with these kind of questions, i reckon they wouldn't ask for the acceleration unless they wanted an equation that looked like mg-T=ma . cos if it's all in equilibrium, it's quite simple as then T=mg, and there's no a anyway.
You've reallllllllly got to start typing "because" rather than "cos" when answering Maths questions. I read your post thinking: "Oh lord, why is she taking components".

(Original post by Coolraj)
i)f = ma
0.5g = 2*a
a = 0.5g/2
ii)0.5g-T=0.5*0.5g/2
make T the subject and you should get the answer! Hope that helped!
I don't think you can say that resultant force acting on the 2kg block is 0.5g. I don't think think you can do this because the force you're talking about (the weight of the 0.5kg block) is acting perpendicular to where the 2kg block would be moving. However, since the pulley is probably smooth I would have thought you'd have just been able to assume all motion to be in a straight line where the second block has a driving force of 0.5g... ?

i.
N2L (2kg) --> T = 2a
N2L (0.5kg) --> 0.5g - T = 0.5a

Bit of manipulation and a = g/5 = 1.96 ms^-2

ii.
T = 2a = 2*1.96 = 3.92 N
0
12 years ago
#7
Here you go :

A block of mass 2kg slides on smooth horizontal table. It is attached to hanging object mass 0.5kg by inelastic string that presses over pulley. Block is released;

i) calculate acceleration of block ii) tension in string

i) T=2a
0.5g - T = 0.5a
0.5g - 2a = 0.5a
2.5a = 4.9
a = 1.96

ii) T=2a
T= 2*1.96
T= 3.92N

Michael
0
#8
Thanks for the clear up guys!

tis helped!
0
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