How can I solve the following problem?

A block that moves at the speed of 120 cm / s keep going for 70 cm before stopping. What's the coefficient of friction between block and table calculated from the deceleration? Perform the same calculation applying the conservation of energy.

I've tried to calculate the time of deceleration with:

$$ \begin{cases} 1.2 = at\\ 0.7 = 1.2t-\displaystyle\frac{1}{2}1.2t \end{cases} $$

And the deceleration with:

$$ 0.7 = 1.2t - \frac{1}{2}at^2 $$

And I got:

$$ \begin{aligned} a~&= 1.0285714285714287\frac{\mathrm{m}}{\mathrm{s^2}}\\ t~&= 1.1666666666666667\mathrm{~s} \end{aligned} $$

But I don't know how to use them to calculate the coefficient of friction with and/or without conservation of energy.

  • $\begingroup$ Acceleration due to friction is negative coefficient of friction times the normal force, and of course the normal force is equal and opposite to the gravitational force, which is the mass times g. Do you have the mass? $\endgroup$ – raptortech97 Sep 15 '14 at 21:47
  • 1
    $\begingroup$ The mass is not needed to solve this problem. $\endgroup$ – Bryson S. Sep 15 '14 at 21:56

No need to involve time in your calculations. If the acceleration is constant, then we know purely from kinematics that $$V_f^2=V_0^2+2a\Delta x,$$ and in this case the final velocity $V_f=0$, so the magnitude of the acceleration is simply $$a=\frac{V_0^2}{2\Delta x}=\frac{F_f}{m}.$$ From the definition of the coefficient of kinetic friction $\mu_k$, we know that $$F_f=\mu_kF_N=\mu_kmg,$$ so $$\mu_k=\frac{F_f}{mg}=\boxed{\frac{V_0^2}{2g\Delta x}}.$$ For the conservation of energy formulation, We know that the kinetic energy lost by the block must be equal to the work done by friction. Mathematically, this is $$\frac{1}{2}mV_0^2=\mu_kmg\Delta x.$$ Cancelling like terms and rearranging, we will again get $$\mu_k=\frac{V_0^2}{2g\Delta x}.$$


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