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Jazz.fm · Oct 29, 2010, 05:53 AM

The question states 1) as an object of mass 1kg is attached to a trolley of mass 2.5kg as shown. Initially the 1kg mass is held at 1.6m above the floor. Ignore the efforts of friction and air resistance.

a) show that when the 1kg mass is released the acceleration of both the trolley and the mass is 2.8ms^-2
am I overcomplecating this and is it just f=ma thanks x

ebaines · Oct 29, 2010, 06:14 AM

Yes, it's just ΣF=ma. Here ΣF = the weight of the 1 Kg block, and m is the total mass of the system, which is 3.5 Kg. So

$ F = ma \\ 1 Kg \times 9.8 \frac m {s^2} = 3.5Kg \times a $

Jazz.fm · Oct 29, 2010, 06:30 AM

thank you I have got the answer 2.8ms^-2

Unknown008 · Oct 29, 2010, 07:40 AM

That is what you had to show :)

Jazz.fm · Oct 29, 2010, 08:44 AM

Originally Posted by ebaines

Yes, it's just ΣF=ma. Here ΣF = the weight of the 1 Kg block, and m is the total mass of the system, which is 3.5 Kg. So

$ F = ma \\ 1 Kg \times 9.8 \frac m {s^2} = 3.5Kg \times a $

How would the acceleration and max velocity change if the height changed from 1.6m to 1m

and also does 1s look right for the amount of time it takes to reach the floor? Thanks x

Unknown008 · Oct 29, 2010, 09:02 AM

You can find the max velocity using the principle of conservation of energy.

The potential energy at the height h is given by:

$E_p = mgh$

The max speed v is obtained when all the potential energy is converted to kinetic energy, or:

[math[E_k = E_p[/math]

$\frac12 mv^2 = mgh$

You can find the max speed from there.

You'll also see that when h is increased, the max velocity is increased.

For the time to reach the ground, use the formula:

$h = \frac12 at^2$