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Chapter 4: Problem 6

\(\mathrm{A}\) high school experiment consists of a block of mass \(2 \mathrm{kg}\) sliding across a surface (coefficient of friction \(\mu=0.6\) ). If it is given an initial velocity of \(5 \mathrm{m} / \mathrm{s}\), how far will it slide, and how long will it take to come to rest? The surface is now roughened a little, so with the same initial speed it travels a distance of \(2 \mathrm{m}\). What is the new coefficient of friction, and how long does it now slide?

Short Answer

Expert verified
The block will slide 2.12m and it will take 0.85s to come to rest. The new coefficient of friction is 0.64 and the block will now slide for 0.8s before coming to rest.

Step by step solution

01

Identify the Given Values and Formula

Given that: mass \(m = 2 \, \mathrm{kg}\), initial velocity \(v = 5 \, \mathrm{m/s}\), coefficient of friction \(\mu = 0.6\). We have our friction force \(F = \mu m g\) where \(g\) is the acceleration due to gravity, and from that we get our acceleration \(a = F/m\). We also use the equation \(v^2 = u^2 + 2as\) to find the distance.
02

Calculate the Frictional Force and the Acceleration

The frictional force \(F = \mu m g = 0.6 \times 2 \times 9.8 = 11.76 \, \mathrm{N}\). The acceleration of the block \(a = F/m = 11.76/2 = 5.88 \, \mathrm{m/s}^2\). Note that the velocity of the block becomes zero when it comes to rest, thus plugging these values into the second equation of motion allows us to solve for distance \(s\).
03

Calculate the Distance and Time

Applying the equation \(0 = v^2 + 2as \Rightarrow s = - v^2 / 2a = -25 / -11.76 = 2.12 \, \mathrm{m}\). The time it takes to come to rest can be calculated by using the equation \( v = u + at \Rightarrow t = (v - u) / a = 0 \), which gives \( t = 5 / 5.88 = 0.85 \, \mathrm{s}\).
04

Identify the New Conditions and Repeat the Calculation

For the second part of the problem, with the roughened surface, the distance the block slides is reduced to \(2 \, \mathrm{m}\). We now need to find the new coefficient of friction \(\mu'\). The initial velocity and the mass of the block remains the same, so we have to find the new acceleration and the coefficient of friction. Applying the same equation for the distance, but this time with the distance \(d = 2m\), we can find the new acceleration. From there, we can get the new coefficient of friction.
05

Calculate the New Coefficient of Friction and Time

From the formula \(s = - (v^2 / 2a') \Rightarrow a' = - v^2 / 2s = -25 / -4 = 6.25 \, \mathrm{m/s}^2\). The new frictional force is \(F' = ma'= 2 \times 6.25 = 12.5 \, \mathrm{N}\). Therefore, the new coefficient of friction is \(\mu' = F' / mg = 12.5 / (2 \times 9.8) = 0.64\). Using the same formula for time as in step 3, but with \(a'\), we get \( t' = 5 / 6.25 = 0.8 \, \mathrm{s}\). Which is the new time the block slides before it comes to rest.

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