Question

A 0.2 kg object is released from rest at point A on a track that is one-quarter of a circle with radius 1.6 m, as shown in the figure. It slides down and reaches point B with a speed of 4.8 m/s. Then it slides on a level surface a distance of 3 m until it comes to rest at point C.
a) How much work is done by friction on the package at it slides down the circular track from A to B?
b) What is the coefficient of kinetic friction on the horizontal surface?
c) What average power is produced by friction as the object moved from B to C?
d) What instantaneous power is produced by friction at point B?

Answer 1

Ýí ÍÇáÉ ÚÏã ÇáÇÍÊßÇß ÓÊßæä : ÇáØÇÞÉ ÇáßÇãäÉ Ýí ÇáÃÚáì = ÇáØÇÞÉ ÇáÍÑßíÉ Ýí ÇáÃÓÝá ( äåÇíÉ ÇáãäÍÏÑ )
æãÚ æÌæÏ ÇáÇÍÊßÇß Óíßæä ÇáÔÛá åæ ÇáÝÑÞ Èíä ÇáØÇÞÊíä ßÇáÊÇáí :
W=mgh-1/2 mv^2=0.2×9.8×1.6-1/2×0.2×〖4.8〗^2=0.832 J

2.b
áÍÓÇÈ ãÚÇãá ÇáÇÍÊßÇß ÇáÍÑßí äÚÊãÏ Úáì ÇáÌÒÁ Èíä B æ C ßãÇ Ýí ÇáÔßá ÇáãÌÇæÑ ÝÝí åÐå ÇáãäØÞÉ
F_Fric=μ F_n but F_n=mg
F_Fric=μ mg
And
F=ma
a=F/m=(μ mg)/m=μ g

μ=a/g……………..1
So
a=v^2/d=〖4.8〗^2/3=7.68 m/s^2
From ..1
μ=a/g=7.68/9.8=0.78


c.
P ̅=W/t=(F_Fric×d)/t=F_Fric×v ̅

but F_Fric=μ mg=0.78×0.2×9.8=1.53N and v ̅=4.8/2=2.4 m/s

∴ P ̅=F_Fric×v ̅=1.53×2.4=3.67 Watt

d.
P=F_Fric×v=1.53×4.8=7.34 Watt