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Question: The maximum frictional force on a motor bike of \(2.5{\text{ }}H.P\) is \(1000{\text{ }}N\). What is...

The maximum frictional force on a motor bike of 2.5 H.P2.5{\text{ }}H.P is 1000 N1000{\text{ }}N. What is the maximum possible speed of the bike?

Explanation

Solution

We know that Power is defined as the work done per unit of time. In this problem we first change the value of power from horsepower to watt as per conversion.The SI unit of Power is Watt. The industrial unit is “horse power, hp”. So, 1 hp=746 watts1{\text{ }}hp = 746{\text{ }}watts.

Complete step by step answer:
As given in problem:
P=2.5 HPP = 2.5{\text{ }}HP
Change the unit horsepower into Watt
P=2.5 ×746 WP = 2.5{\text{ }} \times {\text{746 W}}
And, F=1000 NF = 1000{\text{ }}N
We have to find the maximum possible speed of the bike.
Power=Force×DisplacementTime\text{Power} = \text{Force} \times \dfrac{\text{Displacement}}{\text{Time}}
As, DisplacementTime=velocity\dfrac{\text{Displacement}}{\text{Time}} = \text{velocity}

So,
Power=Force× Velocity\text{Power} = \text{Force} \times {\text{ Velocity}}
We can write,
Velocity=PowerForce\text{Velocity} = \dfrac{\text{Power}}{\text{Force}}
Put the values in above equation
Velocity=2.5×7461000\text{Velocity} = \dfrac{{2.5 \times 746}}{{1000}}
Velocity=18651000\Rightarrow \text{Velocity} = \dfrac{{1865}}{{1000}}
Velocity=18651000\Rightarrow \text{Velocity} = \dfrac{{1865}}{{1000}}
By solving, we get
Velocity=1.865 ms1\therefore \text{Velocity} = 1.865{\text{ }}m{s^{ - 1}}

Therefore, the maximum possible speed of the bike is 1.865 ms11.865{\text{ }}m{s^{ - 1}}.

Note: In order to solve this question we have assumed there will be no loss of power of the motor bike. If we go practically then there will be loss of some part of power in the form of heat due to frictional force and hence we cannot apply the above method to find the maximum velocity of the bike.