Question

A mass of 6 kg is placed on an incline with a coefficient of static friction of 1.80 and coefficient of kinetic friction of 1.28. No additional forces are acting on it and it is raised slowly from a small angle and it begins to slide at some angle. If 10 degrees is added to this angle, and the mass started from rest from a vertical height of 28 meters, how much longer would it take to travel down the incline to the bottom compared to if there was no friction on the incline? Answer in seconds.

Answer 1

mass of the object is acting vertically downwards whicha can be split into two

mgsin acting downwards along the inclined plane and mgcos acting downwards perpendicular to the inclined plane

At certain angle the mass starts to slide down , since there are no other external forces, at that time the force mgsin will be equal to frictional force mgcos

mgsin =mgcos

tan = = 1.8

= tan^-11.8 = 60.95 deg

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now 10 deg is added to the angle

so =70.95 deg

find length of inclined plane

vertical height = 28m , find length of inclined plane

sin = opposite side/hypotenuse = 28/hypo

length = 28/sin70.95

length = 29.62 m

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if there was friction

Fnet = mass* a

mgsin70.95 -mgcos70.95 = 6 *a

mg(sin70.95 - 1.28cos70.95)/6 = a

a =5.17 m/s²

s =ut +1/2at² where s =29.62 m ,u =0 and a =5.17 m/s²

29.62 = 0 +0.5*5.17*t1²

t1= 3.385 sec

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if there was NO friction

Fnet = mass* a

mgsin70.95 = 6 *a

mg(sin70.95)/6 = a

a =9.27 m/s²

s =ut +1/2at² where s =29.62 m ,u =0 and a =9.27 m/s²

29.62 = 0 +0.5*9.27*t2²

t2= 2.53 sec

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Time = t1-t2 = 3.385 - 2.53 = 0.855 sec

ANSWER = It would take 0.855 sec longer to slide down the inclined plane when there is friction than when there is no friction