Satarting from rest and a height of 10 m, a 2-kg mass slidesdown a 30 degree inc

Question

Satarting from rest and a height of 10 m, a 2-kg mass slidesdown a 30 degree incline, reaching the bottom with a speed of 10m/s.What is the approximate work done by friction? [a]0 [b]-103 [c]-100J [d]-200J [e]cannot be determined without the coefficient offriction Satarting from rest and a height of 10 m, a 2-kg mass slidesdown a 30 degree incline, reaching the bottom with a speed of 10m/s.What is the approximate work done by friction? [a]0 [b]-103 [c]-100J [d]-200J [e]cannot be determined without the coefficient offriction

Explanation / Answer

The approximate work done by friction is         Wf = mgd sin30 -(1/2)mv2 Here m = 2kg            g=9.8 m/s2            d = (10m)/sin30            v= 10 m/s On substitution we get           Wf = 100 J            d = (10m)/sin30            v= 10 m/s On substitution we get           Wf = 100 J

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