I'm working through a problem and I get two different answers, one twice the other, depending on which method I use. I have a mass of $0.5$ Kg on a smooth plan (no friction). The plane is inclined at $30^{\circ}$ to the horizontal. The particle starts at rest at point $A$, and slides down the plane under the influence of gravity until it reaches point $B$. When it arrives at point $B$ it is travelling at $6$ m/s. The question is: how far is it from $A$ to $B$?


Gravity pulls down on the particle with a force of $0.5g$ N, where $g \approx 9.8$. This force can be resolved into a component parallel to the slope and a component at right-angles to the slope. The parallel component points down the slope and has length $0.5g\sin(30^{\circ}) = 0.25g$. Thus, the acceleration down the slope is $0.25g$ m/s/s. Using the formula $v=u+at$ we can find the time taken to arrive at $B$. We have $6 = 0 + 0.25gt$ and thus $t= \frac{24}{g}$ s. We can use this value of $t$ and put it into the formula $s = ut + \frac{1}{2}at^2$ which gives $s = 0 +\frac{72}{g} \approx 7.35$ m.

Work-Energy Principle

Because the only work being done is by gravity, we know that the loss of potential energy must equal the gain in kinetic energy. If the particle slides $x$ m down the slope it will have fallen $x \sin (30^{\circ})$ m vertically. It follows that $\Delta E_p = mg\Delta h = 0.5 \times g \times x\sin(30^{\circ})$ J. The partical accelerates from $0$ m/s to $6$ m/s and so $\Delta E_k = \frac{1}{2}m(v^2-u^2) = 0.5 \times 0.5 \times (6^2-0^2) = 9$ J. Putting this together: $\Delta E_p = \Delta E_k \implies x \approx 3.67$ m.

SUVAT twice W-E-Principle

Using the exact values, by first answer is twice the second. The book I'm reading used the Work-Energy Principle and arrives at the answer $x \approx 3.67$ m. I tried to verify the result by using the SUVAT equations, but sadly came up with a different number. I would be most grateful if you could point out my error(s).

  • $\begingroup$ What is "SUVAT" ? $\endgroup$ Dec 24 '12 at 17:17
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    $\begingroup$ A well-known, at least in the UK, mnemonic for the equations of motion. S stands for displacement, U for initial velocity, V for current/final velocity, A for acceleration and T for time. en.wikipedia.org/wiki/SUVAT $\endgroup$ Dec 24 '12 at 17:27
  • $\begingroup$ Thanks for the explanation, Fly by Night. $\endgroup$ Dec 24 '12 at 17:50

I think you are confusing acceleration with force in the "SUVAT" approach.

You calculated the force parallel to the slope correctly to be $0.25g$.

$F=ma$ applies here so $a=F/m=0.25g/0.5=0.5g$

$v=u+at\implies t=\frac{v}{a}=\frac{12}{g}$

So $s=ut+\frac{1}{2}at^2=\frac{1}{2}\frac{g}{2}(\frac{12}{g})^2\approx3.67\textrm { m}$

I hope this is clear enough, else just post a comment to say what I should tell more clearly.

  • $\begingroup$ +1 How very foolish of me. Thanks for that mr. FS. $\endgroup$ Dec 24 '12 at 17:30

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