Recent content by Argonaut

Yes and they give the book solution. Got it. Thanks, both. It makes more sense now that I typed it up and pondered some more.

Here is my attempt at the solution: a) The apparatus may only experience acceleration ##a > g## while in contact with the spring. Since the spring exerts the greatest force when it is the most compressed, the apparatus will undergo the greatest acceleration at that point. So Newton's second...

Thanks! This instantly made me look forward to Abstract Algebra more.

Thanks, @topsquark, @pasmith and @fresh_42, for the insightful replies. I'm much obliged! I now realise that the long division of polynomials has been a glaring blind spot in my mathematical repertoire. (An amusing aside: I blame the inconsistency of mathematical notation across nations...

This is part of a longer exercise I struggled with. I checked the solutions manual, and there was a bit where they performed the following steps: $$x^3=3a^2x-2a^3 \\$$ $$(x-a)^2(x+2a)=0$$ And then concluded that the roots were ##a## and ##-2a##, which is clear. What I can't work out is how...

Thanks, both!

Is my solution correct? (I only have answers to odd-numbered exercises.) Is it a good solution or have I overcomplicated things? (a) The forward force provided by the engine balances the air resistance force, so ##F_{engine}=F_{air} = \alpha v^2 + \beta /v{^2}##. Let ##W_{engine}## be the...

Thanks for the tip. I think I've got it now - at least I've managed to reproduce the correct value. So on the one hand ##W_{tot} = \Delta K = K_f - K_i = 0 - \frac{1}{2}(m_1+m_2)v_0^2##. On the other hand, treating the blocks, the rope and the pulley as a system, there are two forces doing...

No worries, and thanks for the help. Looking forward to my lunch break to have another go at this exercise :biggrin:

It does not, but its descent is 'slowing', so there is negative acceleration (supposing the positive direction is the direction of displacement). I thought that would mean that the force acting in the negative direction is greater than the one acting in the positive direction.

Right, so since there is acceleration, Newton's First Law does not apply, and ##T \neq m_1g ##, in fact ## T > m_1g ##, is that correct? In which case, I'm probably better off considering the whole system instead of block 2.

My final answer is different from the official one in the back of the book, and I can't figure out what I did wrong. This is my attempt: Let block 1 be the vertically moving block and let block 2 be the horizontally moving one. Also, let ##m_1 = 6.00 ~\rm{kg}##, ##m_2 = 8.00 ~\rm{kg}##, ##v_0...

Thank you. This is illuminating.

I see, thanks @Steve4Physics . Potential energy is the next chapter so I'll find out soon enough. And thanks everyone for gently guiding me to the solution. What a great learning experience!

And then getting the vertical component of the displacement ##d\sin{(25°)}=7.35~\rm{m}##. So the ##\sin{(25°)}## term just cancels out. Could I have derived ##y=\frac{(12.0~\rm{m/s})^2}{2(9.80~\rm{m/s^2})}## more easily somehow? Was trigonometry unnecessary to begin with?