Question

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant but starts at .100 at P and increases linerly with distance past P, reaching a value of .600 at 12.5 m past point P. (a) Use the work energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid iff the friciton coefficient didn't increase, but instead had the constant value of .1?

Answer 1

Answer:

a) 5.11 m

b) 0.3

c) 10.33 m

Explanation:

a) Since the friction coefficient is linearly proportional to distance, its formula is given by:

$\mu(x)=Ax+B(1)$

Recall that $\mu(0)=0.1$ and  $\mu(12.5)=0.6$:

$\mu(0)=A(0)+B=0.1\\B=0.1\\\mu(12.5)=A(12.5m)+0.1=0.6\\\\A=\frac{0.6-0.1}{12.5m}=0.04m^{-1}$

Replacing A and B in (1):

$\mu(x)=0.04m^{-1}x+0.1(2)$

According to Newton's second law:

$\sum F_y: N-mg=0\\N=mg\\F_{x}=-f=-\mu N=-\mu mg=-(0.04m^{-1}x+0.1) mg$

The work done by the friction force is:

$W=\int\limits^x_0 {F_{x}} \, dx\\W=\int\limits^x_0 {-(0.04m^{-1}x+0.1) mg} \, dx\\W=-mg[0.04m^{-1}\frac{x^2}{2}+0.1x](3)$

Work-energy principle states that work is the difference of kinetic energy:

$W=\Delta K\\W=K_f-K_i\\W=\frac{m(v_f)^2}{2}-\frac{m(v_i)^2}{2}(4)$

Recall that the final speed is zero. Replacing (3) in (4):

$-mg(0.02m^{-1}x^2+0.1x)=\frac{mv_i^2}{2}\\-9.8\frac{m}{s^2}(0.02m^{-1}x^2+0.1x)=-\frac{(4.5\frac{m}{s})^2}{2}\\0.02m^{-1}x^2+0.1x=\frac{20.25\frac{m^2}{s^2}}{2*9.8\frac{m}{s^2}}\\0.02m^{-1}x^2+0.1x-1.033m=0$

Solving x:

$x_1=-10.11m\\x_2=5.11m$

We take the positive root, so the box slides 5.11 m before stopping.

b) We replace 5.11 m in (2):

$\mu(5.11)=0.04m^{-1}(5.11m)+0.1\\\mu(5.11)=0.30$

c) Using (3) and (4) with constant friction coefficient:

$W=\int\limits F_x\, dx \\W=\int\limits -\mu mg\, dx \\W=-\mu mg \int\limits\, dx \\W=-\mu mgx\\W=-\frac{mv_i^2}{2}\\-\mu mgx=-\frac{mv_i^2}{2}\\x=\frac{v_i^2}{2\mu g}=\frac{(4.5\frac{m}{s})^2}{2*0.1(9.8\frac{m}{s^2}}\\x=10.33m$

Answer 2

The distance to which the box slides before stopping is 5.11 m

The coefficient of friction at the stopping point is 0.3

The distance to which the box would have slid off if the friction coefficient didn't increase, but instead had the constant value of .1 is 10.33 m

Calculation and Parameters:

The formula for the friction coefficient which is linearly proportional to distance is:
u(x)= 0.04m^-1 x + 0.1(2)

To find the work done using the 2nd law of Newton:

W= -mg[0.04m^-1 (x^2)/2 + 0.1 x]

Since the final speed is zero, hence the positive value of x would be:

x= 5.11m

The coefficient of friction would be:
u(5.11) = 0.04 m^-1 (5.11m) +0.1

u(5.11)= 0.30

With constant friction coefficient:

x= 10.33m.

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