Question

A body of mass \[60\text{ }kg\] is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic friction \[0.5\] and \[0.4\] respectively. On applying the same force what is the acceleration?
A) \[0.98m/{{s}^{2}}\]
B) \[9.8m/{{s}^{2}}\]
C) \[0.54m/{{s}^{2}}\]
D) \[5.292m/{{s}^{2}}\]

Answer 1

Hint: A body kept on a rough surface has a certain mass, the friction provided by the rough surface during kinetic and kinetic state are also given. To find the acceleration we use the frictional force method where we subtract the force under friction’s influence for both kinetic and kinetic friction and after that we divide the force by the mass of the body to find the acceleration of the body under such conditions. The formula for the frictional force is:
\[F={{\mu }_{x}}mg,x\in s,k\]
And the formula for the force applied on the body is:
\[F=ma\]
where m is the mass of the body, a is the acceleration of the body, \[{{\mu }_{x}}\] is the frictional force constant with s and k describing kinetic and kinetic friction with g as gravity of \[9.8m{{s}^{-2}}\].

Complete step by step solution:
For Static Friction
When we use the frictional force formula, we take the value of static friction as \[{{\mu }_{k}}=0.5\].
The acceleration of the body dragged is taken as acceleration due to gravity g of \[9.8m{{s}^{-2}}\].
Now placing the values in the formula with mass as \[60\text{ }kg\], we get the force \[{{F}_{s}}\] as:
\[{{F}_{s}}={{\mu }_{k}}mg\]
\[\Rightarrow {{F}_{s}}=0.5\times 60\times 9.8\]
\[\Rightarrow {{F}_{s}}=294N\]
The frictional force for 0.5 static friction is \[294N\].
For Kinetic Friction
When we use the frictional force formula, we take the value of kinetic friction as \[{{\mu }_{s}}=0.4\].
The acceleration of the body dragged is taken as acceleration due to gravity g of \[9.8m{{s}^{-2}}\].
Now placing the values in the formula with mass as \[60\text{ }kg\], we get the force \[{{F}_{k}}\] as:
\[{{F}_{k}}={{\mu }_{s}}mg\]
\[\Rightarrow {{F}_{k}}=0.4\times 60\times 9.8\]
\[\Rightarrow {{F}_{k}}=235N\]
Hence, the frictional force for 0.4 kinetic friction is \[235N\].
Now we subtract the frictional forces of kinetic and static friction and find the difference in the force when kept in different states.
The difference in the force from both the frictions as:
\[\Rightarrow {{F}_{k}}-{{F}_{s}}\]
\[\Rightarrow {{F}_{k}}-{{F}_{s}}=294N-235N\]
\[\Rightarrow {{F}_{k}}-{{F}_{s}}=59N\]
After finding the difference in the frictional force, we equate the difference in the frictional force by the force of the object without friction coefficient, we get the force F as:
\[F=ma\]
\[F=60\times a\]
Placing the value of the F as the difference in the force obtained from the static and kinetic friction, we get the value of the acceleration as:
\[59=60\times a\]
\[\Rightarrow 59=60\times a\]
\[\Rightarrow a=\dfrac{59}{60}\]
\[a=0.98m{{s}^{-2}}\]

Therefore, the acceleration formed on the body due to the body’s motion without friction is \[0.98m{{s}^{-2}}\].

Note: The difference between the static and kinetic friction is that in static friction the frictional force is greater than that of the applied force on the object and kinetic frictional force is when the applied force is greater than the frictional force present on the body and the difference between these two types of frictional forces is the force needed to move the body from static to kinetic.