Question

In: Physics

A piece of luggage is being loaded onto an airplane by way of an inclined conveyor...

A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 25.0 kg,25.0 kg, travels ?=6.50 md=6.50 m up the conveyor belt at a constant speed without slipping.

If the conveyor belt is inclined at an angle ?=50.0∘,θ=50.0∘, calculate the work done on the bag by

  • the force of gravity (?g)(Wg)
  • the normal force (?N)(WN)
  • the friction force (?f)(Wf)
  • the conveyor belt (?conveyor)(Wconveyor)
  • the net force (?net)(Wnet)

Match the work done by each force to the appropriate energy value.

1590 J1590 J1220 J1220 J−1590 J−1590 J−1220 J−1220 J0 J0 J1020 J1020 J−1020 J−1020 J

?fWf

?conveyorWconveyor

?gWg

?NWN

?netWnet

Answer Bank

Solutions

Expert Solution

Part a.

Work-done by gravity

Since gravity force is always downward and displacement is on inclined plane, So

= Angle between gravity force and displacement = (90.0 + 50.0) deg = 140.0 deg

So,

Wg = Fg*d*cos 140.0 deg

here, Fg = m*g = (25.0 kg)*(9.81 m/s^2) = 245.25 N

d = displacement = 6.50 m

Wg = 245.25*6.5*cos(140 deg) = -1220 J

Part b.

Work-done by Normal force

Since Normal force is always perpendicular and displacement is towards inclined plane, So

= Angle between Normal force and displacement = 90 deg

So,

Wn = N*6.5*cos 90 deg

Wn = 0 J = Work-done by normal force

Part c.

Work-done by Friction force

Since friction force will be in the same direction in which conveyor belt is moving, So that piece of luggage remains stationary on the belt. So

= Angle between Friction force and displacement = 0 deg

So,

Wf = Ff*d*cos 0 deg = Ff*d

here, Ff = friction force

Since luggage is moving at constant speed.

So, by force balance,

Ff = m*g*sin A

given, A = inclined angle = 50.0 deg

then, Ff = 25.0*9.81*sin(50.0 deg)

Wf = Ff*d = 25.0*9.81*6.50*sin(50.0 deg)

Wf = 1220 J = Work-done by friction force

Part d.

Work-done by conveyor belt = Work-done by friction force

So,

W_conveyor = 1220 J

Part E.

Net work-done on the box will be:

W_net = Wg + Wn + Wf

W_net = -1220 + 0 + 1220

W_net = 0 J

Let me know if you've any query.


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