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Apologies if the following problem does not seem to involve potential energy, it is a homework problem that is labeled under the potential energy chapter.

Question

While a roofer is working on a roof that slants at 39.0∘ above the horizontal, he accidentally nudges his 95.0N toolbox, causing it to start sliding downward, starting from rest. If it starts 5.00m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 19.0N ?

Approach

Find $\mu_k,$ Assume $g=9.81\frac{m}{s^2}$

$95\mu_kcos(39^o)=19$ //assumes a rotated coordinate system

$\mu_k=.155$

$\Delta F=mgsin(39^o)-\mu_kmgcos(39^o)=ma$ //m's cancel

$a=gsin(39^o)-\mu_kgcos(39^o)=4.99$

$v^2=v^2_0+2a(x-x_o)$ //v0, x0 = 0

$v=\sqrt{2(4.99)(5)}=7.06$

My approach doesn't use potential energy in any way yet i don't see how it could be wrong. Also, if someone could walk me through an approach using potential energy if it's possible to do it that way, it'd be greatly appreciated

Approach 2

$U_g=K_f$

$mgh = \frac{mv^2}{2}-F_fd$

find h

$sin(39)=\frac{h}{5} => h = 3.15$

$v=\sqrt{\frac{2(mgh+F_fd)}{m}}$

$v=\sqrt{\frac{2((95)(3.15)+(19)(5))}{95}}=2.69$//still wrong answer

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  • $\begingroup$ In your Approach 2, the force value 95 N is being used where a mass value 'm' is expected (in the denominator) $\endgroup$
    – Ryan Cavanaugh
    Commented Feb 19, 2014 at 23:57

1 Answer 1

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First of all don't insert actual numbers until the end. It makes it much easier to keep track and check whether your units check out.

This problem is easier if you invoke conservation of energy. Simply equate:

At t = 0.

-Potential gravitational energy.

At the end.

-Kinetic energy of the toolbox.

-Dissipated energy due to friction.

You will find that these are equal.

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  • $\begingroup$ So I'm assuming you are using the conservation of mechanical energy theorem. I thought you couldn't use it since it is only valid for systems with conservative forces, and I assumed friction was not one. Also, could you mind pointing out where in my logic I went wrong, or is it impossible to do it this way. If so, why? $\endgroup$
    – sreya
    Commented Feb 19, 2014 at 23:21
  • $\begingroup$ It's hard to follow it when you don't keep track properly. Also, energy is ALWAYS conserved - as long as you can properly account for the frictional term(s) feel free to use it. $\endgroup$
    – Kvothe
    Commented Feb 19, 2014 at 23:41
  • $\begingroup$ Do I not keep track properly, seems pretty easy to read to me. I tried doing it the other way but still arrived at a wrong answer, care correcting me one last time, I really appreciate it $\endgroup$
    – sreya
    Commented Feb 19, 2014 at 23:46
  • $\begingroup$ Its mass is not 95 - it's gravitational force is 95N. Furthermore the friction term is positive - after all, it makes you end up with energy (heat). Always keep checking - vary parameters in your head and check whether it makes sense - higher friction shouldn't make the box go faster! $\endgroup$
    – Kvothe
    Commented Feb 19, 2014 at 23:54

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