Question
Question: A body is projected along a rough horizontal surface with a velocity of 6 m/s. If the body comes to ...
A body is projected along a rough horizontal surface with a velocity of 6 m/s. If the body comes to rest after travelling a distance of 9 m, the coefficient of sliding friction is (g=10s2m)
A) 0.5
B) 0.6
C)0.4
D)0.2
Solution
Here this type of question friction force is responsible for the body moving on a rough surface and due to this friction velocity of body becomes zero. In such a case, equilibrium is established between applied force and friction force and the body comes to the state of rest. Using this logic will help you to find the required data.
Complete step-by-step solution:
Since the body is moving along a rough horizontal surface then due to the roughness of the surface, friction force acts and due to this friction, velocity of body decreases continuously and finally after some time the body comes to rest. This logic is followed in this question.
So the body is moving with a velocity of 6m/s.
Let us assume it as initial velocity of the body as represented by u.
u = 6m/s
Due to motion along rough horizontal surfaces the body comes to rest.
Let us assume its final velocity as v.
So, v=0 m/s.
The displacement covered by the body is 9m.
Let us assume this displacement is as S
So, S = 9m
Apply third equation of motion so that acceleration can be obtained
v2=u2+2aS
⇒(0)2=(6)2+2a(9)
⇒0=36+18a
∴a=−2s2m
Negative acceleration means velocity is decreasing and friction is responsible for this decrement.
Due to friction, the body comes to rest.
So we apply,
Applied Force = Friction Force --- (Equation 1)
Applied Force = mass x acceleration.
Friction force = Coefficient of friction x Reaction force
Apply this in Equation 1 we get
Mass x Acceleration = Coefficient of friction x Reaction force.
This above equation can be mathematically written as,
Let us assume mass of body is represented by m , acceleration is represented by a , Coefficient of friction is represented by μ and reaction force is represented by R
⇒m×a=μ×R
Since we know that R = mg
⇒m×a=μ×mg