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A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point

ID: 1784361 • Letter: A

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 0.100 at P and increases linearly with distance past P, reaching a value of 0.600 at 12.5 m past point P. Part A Use the work-energy theorem to find how far this box slides before stopping. m past point P Submit Give Up Part B What is the coefficient of friction at the stopping point? = Submit Give Up Part C How far would the box have slid if the friction coefficient didn't increase, but instead had the constant value of 0.100?

Explanation / Answer

uk = 0.1 + k*x

at x = 12.5 m


uk = 0.6


0.6 = 0.1 + k*12.5

constant k = 0.04


work done dW = fk*dx*cos180 = -fk*dx

W = integration dW

W = integration-uk*m*g*dx


W = integration -m*g*(0.1 + k*x)*dx


W = -m*g*(0.1x + kx^2/2)


work energy relation W = change in KE = Kf - Ki

W = 0 - (1/2)*m*vi^2

-m*g*(0.1x + kx^2/2) = -(1/2)*m*vi^2

9.8(0.1x + 0.04x^2/2) = (1/2)*4.5^2

x = 5.1 m <<<<<--------ANSWER

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at x = 5.1 m

u = 0.1 + (0.04*5.1) = 0.304 <<<------------------ANSWER

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part C

work = -f*x = -u*m*g*x

W = dKE = 0 - (1/2)*m*vi^2

-u*m*g*x = - (1/2)*m*vi^2

0.1*9.8*x = (1/2)*4.5^2


x = 10.33 m <<<------------------ANSWER