We have two a pulley system with two masses of 2kg and 3kg on two ends of a massless string. At the end of 5 seconds, the string breaks. How far up did the 2kg block move?

Here's what I have tried:- $$a_{net}=\frac{F_{net}}{m_1+m_2}$$ $$a_{net}=\frac{3g-2g}{3+2}$$ $$a_{net}=1.96m/s^2$$ $$v=u+at$$ $$v=0+(1.96)(5)$$ $$v=9.8m/s$$ Now the 2kg block moved with velocity 9.8m/s till the string broke, so all the kinetic energy during this travel converted to potential energy

$$PE=\frac{1}{2}mv^2$$ $$PE=96.04J$$ $$mgh=96.04J$$ $$h=4.9m$$

So this happens to be the correct answer but I wanted to use an alternative method which should work but isn't:- $$S=\frac{1}{2}at^2$$ $$S=\frac{1}{2}(1.96)5^2$$ $$S=24.5m$$

Can someone explain why using the basic kinematics equation doesn't work?


1 Answer 1


You are calculating two very different things with your two approaches, possibly because of your false statement: "the 2kg block moved with velocity 9.8m/s till the string broke". That is not true. It was moving at $9.8 m/s$ at the instant the string broke, but was accelerating uniformly to get to that speed for the previous 5 seconds (with an acceleration of g/5, as you correctly worked out).

You first calculation is taking the kinectic energy of the 2kg block at the moment the string breaks, and using energetic arguments, working out how much higher it will go (after the string breaks) above that point (as a projectile starting with an upwards velocity of $9.8 m/s$ moving under the force of gravity).

The second calculation is how far the 2kg block has moved before the string breaks, and is correct.

The initial question as you have presented it is ambiguous as to which answer you are looking for. If it is the distance moved during the first 5 seconds, then you are looking at a block starting from rest and accelerating at $1.96 ms^{-2}$ for 5 seconds, and $24.5 m/s$ is the right answer.

If you are looking for how much higher the 2kg block moves after the string breaks, the you have a block staring with an upward velocity of $9.8 m/s$ and accelerating downwards under gravity at $-9.81 ms^{-2}$ and whether you solve that via kinematic or energetic methods, you will get $4.9 m$.

  • $\begingroup$ Seems like the answer should be 29.4 meters. I read "How far up did the 2kg block move?" to mean how high does the 2kg block go. It travels up 24.5 meters before the string breaks and goes up another 4.9m before freefall. $\endgroup$
    – MaxW
    Oct 29, 2015 at 16:26

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