Question

What is the work done by a person in carrying a suitcase weighing 10 kg f on his head when he travels a distance of 5 m in the vertical direction and horizontal direction

Answer 1

Heya !!

$\text{Here, Since, he is caring the suitcase over his head, force is acting perpendicularly.}\\\\\text{Since, work \: \: done \: (w) \: = \\: Force \: (F) \times \Displacement \: (s) \times \cos(\theta) }\\\\\text{And the force is acting perpendicularly, so, \theta =90^\circ }\\\\\therefore w = F \times s \times \cos(90^\circ)\\\\\implies w = F \times s \times 0 \\\\\because \: \: \cos(90^\circ) \: \:= \: \: 0 \: .... \\\\\textbf{Thus, work done = \underline{\boxed{\bold{0 \: \:Joules}}}}$

Hope it helps !!

Answer 2

Answer:

Work Done = Force × Displacement

If there is an angle change in the force direction, then the formula would be :

Work Done = Force × Displacement × Cos Ф

Now here the suitcase is being carried perpendicular to the ground and displacement is parallel to the ground.

Hence Work Done = Force × Displacement × Cos Ф

=> Work Done = 10 kg f × 5 m × Cos 90

We know that Cos 90 = 0

Hence anything multiplied by 0 is zero.

Hence Work Done = 5 × 10 × 0 = 0 J

Hence there is no Work Done in this case.