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

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.

1)Use the work-energy theorem to find how far this box slides before stopping.

2)What is the coefficient of friction at the stopping point?

3)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

u = 0.1 + kx [ u is coefficient of friction, x is distance from P] determine k: 0.6 = 0.1 + 12.5k k = 0.5/12.5 k = 0.04 u = 0.04x + 0.1 W = K Initial Energy= ½mv² This should be equal to work done against friction Work done against Friction =u*mgdx [ From 0 to x] = mg(0.04x + 0.1)dx =mg(0.02x^2 +.1x) This should be equal to ½mv²=mg(0.02x^2 +.1x) =>0.5*v^2=9.8(0.02x^2+0.1x) [v=4.5] solving for x 1)x=5.109 m[distance it will travel before stopping] 2)At stopping point u= 0.04*5.109 + 0.1=0.30436 3)if u=0.1 0.5*v^2 = 0.1*9.8*x [v=4.5] solving for x x=10.331m

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