Question

Lindsay is standing in the middle of the aisle of a bus that’s at rest at a stop light. The light turns green and the bus speeds up. Without grabbing or hanging on to anything, Lindsay manages to remain stationary with respect to the floor. While the bus is speeding up, the net force on Lindsay points in which direction?
a. Toward the front of the bus b. There is no net force on Lindsay c. Toward the back of the bus

Answer 1

Answer:

b. There is no net force on Lindsay.

Explanation:

Since Lindsay manages to stay stationary then this means that either there is no force acting on Lindsay or all the forces acting on Lindsay are balanced and produces no net force on Lindsay with respect to the bus.

Answer 2

Answer:

No net force acting on Lindsay.              

Explanation:

Solution:-

- Lets assume the mass of the bus to be = M

- Assume mass of Lindsay = m

- The initial speed of bus and lindsay was vi = 0 m/s

- We will consider the system ( Lindsay + Bus ) to be isolated with no fictitious unbalanced forces acting on the system.

- Since, the system can be isolated the application of principle of conservation of linear momentum is completely valid.

- We will assume the direction of bus moving to be positive.

- The principle of conservation of momentum states:

                                 

                           Pi = Pf

                           vi*(m+M) = m*vL + M*vb

Where,

             vL : The velocity of Lindsay after bus starts moving

             vb : The velocity with which bus moves

Since,    vi = 0 m/s.

                           0 = m*vL + M*vb

                          vL = - ( M / m )*vb

- The lindsay moves in opposite direction of motion of bus. We will apply the Newton's second law of motion on Lindsay:

                           Impulse = F*t = m*( vL - vi )

                           Impulse = F*t = m*( - ( M / m )*vb - 0 )

                           Impulse = F*t = -M*vb

                          F = -M*vb/t

- We see that a force ( F ) acts on Lindsay as soon as the bus starts moving. The negative sign shows the direction of force which is opposite to the motion of bus. So the force acts on Lindsay towards the back of the bus.      

- However, if we consider the system of ( bus + Lindsay ), the net force exerted on Lindsay remains zero as the impulse force ( F ) becomes an internal force of the system and is combated by the friction force ( Ff ) between Lindsay's shoes and the aisle floor.

- The magnitude of Friction force ( Ff ) is equivalent to the impulse created by force ( F ) by Newton's third Law of motion. Every action has its equal but opposite reaction. Hence,

                          Fnet = F + Ff

                          Fnet = F - F

                          Fnet = 0

- However, The force ( F ) created by the rate of change of momentum of bus must be less than Ff which is equivalent to weight of Lindsay:

                          Ff = m*g*us

Where,

             us: The coefficient of static friction.

- So for forces to remain balanced, with no resultant force then:

                         Ff ≥ F

                         m*g*us ≥ M*ab

                         us ≥ (M/m)*(ab/g)

- The coefficient of static friction should be enough to combat the acceleration of bus ( ab ).