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Grade 11Mechanics

A 30kg mass rests on a rough horizontal surface a force 200N is applied on the block the block acquires a speed of 4ms^-1 starting from rest in 2s what is teh value of coefficient friction

Profile image of Simeen
6 Years agoGrade 11
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1 Answer

Profile image of Vikas TU
6 Years ago
Dear student 
Initial velocity u = 0
final velocity = 4 m/s
t = 2
acceleration of the body = a
v-u = at
a = 4/2 = 2m/s^2
net force acting on the body = f = ma = 30(2) = 60 N
applied force = 200 N
friction force = applied force - net force = 200-60 = 140 N
coefficient of friction = frictional force/weight = 140/300 = 0.4 
 
Hope this information will help clear you doubt about force.