In a truck-loading station at a post office, a small 0.200-kg package is released from rest a point A on a track that is one quarter of a circle with radius 1.60 m.The size of the packageis much less than 1.60m, so the package can be treated at aparticle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distanceof 3.00 m to point C, where is comes to rest.&nbsp;(a) What is the coefficient of kinetic friction on the horizontal surface?&nbsp;(b) How much work is done on the package by friction as it slides down the circular arc from A to B?

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In a truck-loading station at a post office, a small 0.200-kg package is released from rest a point A on a track that is one quarter of a circle with radius 1.60 m.The size of the packageis much less than 1.60m, so the package can be treated at aparticle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distanceof 3.00 m to point C, where is comes to rest.&amp;nbsp;(a) What is the coefficient of kinetic friction on the horizontal surface?&amp;nbsp;(b) How much work is done on the package by friction as it slides down the circular arc from A to B?

Elberte · Accepted Answer

a.) The kinetic energy of the package at point determines how much work is done by friction.μkmgL=12mvB2&amp;nbsp;Then solve for&amp;nbsp;μk&amp;nbsp;μk=vB22gL=(4.8)22(9.8)(3)=0.392&amp;nbsp;b)&amp;nbsp;Wother=KB−UA&amp;nbsp;Wother=12mv2−mgh&amp;nbsp;=12(0.2)(4.8)2−(0.2)(9.8)(1.6)&amp;nbsp;=−0.832&amp;nbsp;J

Jeffrey Jordon · Accepted Answer

a) Kinetic friction on horizontal surface is equal to friction force  μNdcos⁡(θ)=12m(v22−v12)  N=mg  Given  m=0.200kg;d=3.00m;v1=4.80m/s;v2=0  Therefore   μ(0.200kg)×9.8×3cos⁡180∘=12(0.200)×(02−4.802)  Step 2  b) Total work= work done by gravity + work done by friction  Total work = kinetic energy  WT=12m(v12−u2)  Wg+Wf=12m(v12−u2)  Wf=12m(v12−u2)−mgR=120.200((4.80)2−0))−(0.200)(9.8)(1.60)  Wf=−0.832J (work done against moving body , hence -ive sign)

Jeffrey Jordon · Accepted Answer

Step 1  Given:  The mass is 2 kg  The radius is 1.60 m  The package&#039;s size is 1.6 m  The speed is given as 4.80 m/s  The distance is 3 m  Step 2  a) At point B, kinetic energy is equal to the friction&#039;s work done. The formula is denoted as,  12mvB2=μkmg×Bc  μk=vB22g×Bc=4.822×9.81×3  μk=0.39  Step 3  b) The sum of kinetic energy at point B and work done by the friction is equal to the Potential energy at point A.  The formula is denoted as,  mgR=WFAB+12mvB2  WFAB=m(gR−12vB2)  On substituting the values,  WFAB=0.2(9.81×1.60−0.5×4.82)  WFAB=0.835J  Step 4  Solution:  a) At the horizontal surface, the coefficient of friction is 0.39  b) From A to B, the friction&#039;s work done is 0.835 J

In a truck-loading station at a post office, a small 0.200-kg package is released from rest a point A on a track that is one quarter of a circle with

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