A waiter is carrying a tray above his head and walking at a constant velocity. If he applies a force of 5.0 newtons on the tray and covers a distance of 10.0 meters, how much is the work being done?

Respuesta :

The correct answer to the question is- zero i.e no work is done by the body.

CALCULATION:

As per the question, the force applied by waiter on tray F = 5.0 N.

The waiter is carrying the tray above his head .

Hence, displacement of waiter = displacement of tray.

Hence, displacement of the tray S = 10.0 m

The angle between force and displacement is 90 degree as the normal force on book is vertically upward and displacement is in horizontal direction.

Hence, amount of work done is calculated as -

                                Work done W = [tex]\vec F.\vec S[/tex]

                                                        = [tex]FScos\theta[/tex]

                                                        = [tex]FScos90[/tex]

                                                        = 5.0 N × 10.0 m × 0

                                                        = 0.

Hence, the work done will be zero.

Answer:

W = 0

Explanation:

It is given that, a waiter is carrying a tray above his head and walking at a constant velocity. He applies a force of 5.0 newtons on the tray and covers a distance of 10.0 meters. We know that the product of force and displacement is called the work done.

The expression for the work done is given by :

[tex]W=Fd\ cos\theta[/tex]

Here, the angle between the force and the displacement is 90 degrees. So,

[tex]W=Fd\ cos(90)[/tex]

W = 0

So, the work done is 0. Hence, this is the required solution.