Question

an object on mass 10kg moving with a velocity of 5.6 m/s on a rough horizontal surface comes to rest after traveling a distance of 16m. find
1. coefficient of friction
2. work done against friction
3. time taken by the body to come to rest

Answer 1

Given:

Initial velocity (u)= 5.6 m/s
v= 0. (Body comes to rest)
mass= 10 kg
Distance(s)= 16m


2as = v²-u² ( III EQ OF MOTION)


2×a×16= (0)² -(5.6)²


32a= -31.36


a= -31.36/32

a = -0.98 m/s²


Force(F) = mass(m) × acceleration (a)

F = ma

F = 10kg ×(-0.98) m/s²


F = -9.8 N

Hence, force of friction = 9.8 N

Work done(W) = Force(F)× Distance(s)

W= F× s

W= -9.8× 16

W = -156.8J

Hence, the work done against friction is 156.8


v= u+at (Ist eq. Of motion)

0= 5.6+(-0.98)t

0= 5.6 -0.98 t

-5.6 = -0.98 t

t = 5.6/0.98

t = 5.71428…

t= 6 sec (approximately

Hence, time taken by the body to come to rest is 6 sec (approximately)

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Hope this will help you....