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Question

Question: If a body of mass m is moving on a rough horizontal surface of coefficient of kinetic friction \[\mu...

If a body of mass m is moving on a rough horizontal surface of coefficient of kinetic friction μ\mu , the net electromagnetic force exerted by surface on the body is
A. mg1+μ2mg\sqrt {1 + {\mu ^2}}
B. μmg\mu mg
C. mgmg
D. mg1μ2mg\sqrt {1 - {\mu ^2}}

Explanation

Solution

The frictional force is the self-adjustable dissipative force that works against the direction of motion of the body. The force of friction is proportional to the normal force exerted by the surface on the body, and the proportionality constant is known as the coefficient of friction, μ\mu .

Complete step by step answer:
According to the question;
Given:
The mass of the body, m.
The normal reaction force provided by the surface on the body, FN=mg{F_N} = mg
The frictional force acting on the body in the direction opposite to the motion of the body is;
Ff=μmg{F_f} = \mu mg
So, the net electromagnetic force exerted by the surface on the body is;

{F_{net}} = \sqrt {{F_N}^2 + {F_f}^2} \\\ = \sqrt {{{\left( {mg} \right)}^2} + {{\left( {\mu mg} \right)}^2}} \\\ = mg\sqrt {\left( {1 + {\mu ^2}} \right)} \end{array}$$ Therefore, the net electromagnetic force exerted by the surface on the body is $$mg\sqrt {\left( {1 + {\mu ^2}} \right)} $$ **So, the correct answer is “Option A”.** **Additional Information:** The frictional force can be of two categories – static and dynamic. The static friction acts on the body when it is at rest; on the other hand, the dynamic friction acts on the body while the body is in motion. The static force of friction is self-adjustable while the peak value of the force of friction that can act on a body in a given situation is the value of the force of friction that acts on the body just at the point when the body starts to move. **Note:** In the current question, the force of friction acts in the direction opposite to that of the motion of the body. On representing the frictional force vectorially, we have to add a negative sign in front of the magnitude of the force of friction ($${\vec F_f} = - \mu mg$$). On calculating the net force, we square and add both the values of the force, hence the negative sign does not appear in the net force.