Question

Reena is riding a motorized scooter along a straight path. The combined mass of Reena and her scooter is 80 kg. The frictional forces that are acting total to 45 N. What is the magnitude of the driving force being provided by the motor if –
(i)She is moving with a constant speed of 10\[m{{s}^{-1}}\].
(ii)She is accelerating at 1.5 \[m{{s}^{-2}}\].

Answer 1

Hint: We need to understand the dependence of the mass, velocity and the acceleration of a moving body and the frictional force acting on the body. We can use this information to solve the problem by finding the magnitude of the force in each case.

Complete step by step answer:
We are given a rider who is riding a motorised scooter against the frictional force acting on her scooter due to the combined mass of Reena and the scooter. We know that the frictional force acting on a body in motion is a constant. The frictional force always provides resistance to the motion throughout the motion.
We are given two situations when the system of motion undergoes different kinds of motion. In the first situation, the system is moving with a constant speed, whereas, in the second situation, the system is moving with an acceleration throughout the motion. We can consider each of the case to analyse the driving force in each case.
(i)The rider and the scooter of mass 80 kg is moving with a constant speed of 10 \[m{{s}^{-1}}\]. We know that an acceleration is required to produce an excess force in the system. Since, the body is moving with a constant speed, we can understand the force acting on the body will be as to balance the frictional force. The driving force is given as –
\[\begin{align}
  & {{F}_{dr}}={{F}_{f}}+F \\
 & \Rightarrow {{F}_{dr}}=45N+ma \\
 & \Rightarrow {{F}_{dr}}=45N+0 \\
 & \therefore {{F}_{dr}}=45N \\
\end{align}\]
The force which drives the system is 45N.
(ii)The rider and the scooter of mass 80 kg is moving with an acceleration of 1.5\[m{{s}^{-2}}\]We can find the driving force as –
\[\begin{align}
  & {{F}_{dr}}={{F}_{f}}+F \\
 & \Rightarrow {{F}_{dr}}=45N+ma \\
 & \Rightarrow {{F}_{dr}}=45N+(80\times 1.5) \\
 & \Rightarrow {{F}_{dr}}=45N+120N \\
 & \therefore {{F}_{dr}}=165N \\
\end{align}\]
The force which drives the system in this case is 165 N.

The solution for the driving forces in the two cases are –
(i)45 N
(ii)165 N

Note: We know that a body can undergo a uniform motion at a constant speed only if there exists no net force in the system. In the first situation of the problem, the frictional force and the driving force act in the opposite directions and cancel off each other.