
A block of metal is lying on the floor of a bus. Find the maximum acceleration which can be given to the bus so that the block may remain at rest.
A) $ \mu g $
B) $ \mu /g $
C) $ {\mu ^2}g $
D) $ g/\mu $
Answer
552.9k+ views
Hint: We assume that the bus is moving on a horizontal plane and that there is a coefficient of friction $ \mu $ between the surface of the bus and the piece of metal. The friction force acting on the object will be proportional to the normal force acting on the block.
Complete step by step solution:
As mentioned in the hint, we assume that there is a coefficient of friction $ \mu $ between the surface of the bus and the piece of metal as for a real bus, there will always be friction on its floor.
The block, assuming it has mass $ m $ , will have weight $ mg $ where $ g $ is the gravitational acceleration acting on the object. Since there is no other force on the block in the vertical direction, the normal force acting on the block by the surface of the bus will be
$ N = mg $
Now we know that the friction force that will act on this block will be
$ {F_s} = \mu N $
Substituting $ N = mg $ in the above equation, we get
$ \Rightarrow {F_s} = \mu mg $
Now for the block to stay at rest, the force exerted on the block due to the acceleration must balance the friction force so that the block doesn’t move due to friction which gives us
$ {F_{max}} = \mu mg $
Since the acceleration with the maximum force is $ {F_{max}} = m{a_{max}} $ , we can write
$ m{a_{max}} = \mu mg $
Dividing both sides by mass, we get
$ {a_{max}} = \mu g $ which corresponds to option (A).
Note:
The assumption that there will be friction between the bus and the metal surface is very important as, if there was no friction, the block would start moving for any non-zero acceleration of the bus. Also since the bus is moving in a horizontal direction, there is no vertical force acting on the metal surface however if it were moving on an inclined plane, we would have to take into account the effect of the inclination.
Complete step by step solution:
As mentioned in the hint, we assume that there is a coefficient of friction $ \mu $ between the surface of the bus and the piece of metal as for a real bus, there will always be friction on its floor.
The block, assuming it has mass $ m $ , will have weight $ mg $ where $ g $ is the gravitational acceleration acting on the object. Since there is no other force on the block in the vertical direction, the normal force acting on the block by the surface of the bus will be
$ N = mg $
Now we know that the friction force that will act on this block will be
$ {F_s} = \mu N $
Substituting $ N = mg $ in the above equation, we get
$ \Rightarrow {F_s} = \mu mg $
Now for the block to stay at rest, the force exerted on the block due to the acceleration must balance the friction force so that the block doesn’t move due to friction which gives us
$ {F_{max}} = \mu mg $
Since the acceleration with the maximum force is $ {F_{max}} = m{a_{max}} $ , we can write
$ m{a_{max}} = \mu mg $
Dividing both sides by mass, we get
$ {a_{max}} = \mu g $ which corresponds to option (A).
Note:
The assumption that there will be friction between the bus and the metal surface is very important as, if there was no friction, the block would start moving for any non-zero acceleration of the bus. Also since the bus is moving in a horizontal direction, there is no vertical force acting on the metal surface however if it were moving on an inclined plane, we would have to take into account the effect of the inclination.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

