Tom Sawyer and Huckleberry Finn have decided to go iceblocking. They want to hav

Question

Tom Sawyer and Huckleberry Finn have decided to go iceblocking. They want to have a race down the grass hill near the river. But first, they need to steal two small blocks of ice from the tavern. Over the next two days they happily devise and stealthily execute their plan to acquire the necessary items. Upon completion of their mission, they find themselves at the top of the chosen hill, ready to race down the steepest part. Huck says that he should get the smoother side of the “race track” since the idea was his in the first place, but Tom quickly buys him off with the white marble and dried up lizard tail in his pocket that he thought to bring with him.

As it turns out, the angle of the slope, ?, and the distance travelled, d, is exactly the same for Tom and Huck. However, it takes Huck 1.5 times as long to reach the finish line. Assuming Tom’s path was frictionless, show that the coefficient of kinetic friction for Huck’s path is given by:

Please explain every part fully and show all steps.

Explanation / Answer

let theta is the angle of inclination.

acceleration of Tom, a_T = g*sin(theta)

let t is the time taken to cross the finish line for Tom.

let d is the distance travelled.

so,

d = u*t + 0.5*a_T*t^2

d = 0 + 0.5*g*sin(theta)*t^2

d = 0.5*g*sin(theta)*t^2 ----(1)

acceleration of Huck, a_H = g*sin(theta) - mue_k*g*cos(theta)

time taken for Huck to cross the line = 1.5*t

d = u*1.5*t + 0.5*a_T*(1.5*t)^2

d = 0 + 0.5*(g*sin(theta) - mue_k*g*cos(theta))*2.25*t^2

d = 0.5*(g*sin(theta) - mue_k*g*cos(theta))*2.25*t^2   --(2)

from equations 1 and 2

0.5*g*sin(theta)*t^2 = 0.5*(g*sin(theta) - mue_k*g*cos(theta))*2.25*t^2

g*sin(theta) = (g*sin(theta) - mue_k*g*cos(theta))*2.25

sin(theta) = (sin(theta) - mue_k*cos(theta))*2.25

2.25*mue_kcos(theta) = sin(theta)*(2.25 - 1)

2.25*mue_k*cos(theta) = 1.25*sin(theta)

mue_k = (1.25/2.25)*tan(theta)

= (125/225)*tan(theta)

= (5/9)*tan(theta) <<<<<<<<<<<-----------ANswer

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