Question

$\huge\underline\mathfrak\red{Question}$

Find the work done by a man in carrying a suitcase of mass 20 kg over his head when he

travels a distance of 15m along i) vertical direction
ii) horizontal direction and
iii) a smooth surface inclined at 30° to the horizontal.

Answer the above question with explaination.. ✍️​

Answer 1

${\huge{\boxed{\mathcal{\red{Answer}}}}}$

I) work = 0

ii) work = 2940 joules

iii) work= 2352joules

${\huge{\boxed{\mathcal{\red{Explanation}}}}}$

We have given that -

Mass of suitcase = 20kg

Distance travelled = 15m.

We have to find -

1) Work done man in vertical direction.

Now work is said to be done when a force is applied on a body and the body undergoes displacement.

In this case work done = potential energy of body. = m × g(acceleration due to gravity) × h or s

After inputting the given values in this equation we get.

Work done in vertical direction = 20kg × 9.8 × 15

$= 196 \times 15$

= 2940J.

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2) Now as we know that the accelerating force acting on suitcase is of gravity.

Force of gravity always acts in vertical direction and the suitcase in this case is moving in horizontal direction.

Since the force in horizontal direction is zero thus the work done in horizontal direction is also 0.

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3) For any horizontal component the work done is the product of force, displacement and cosine of the angle.

Here :

Force = m×a = 20×9.8 = 196N

Displacement = 15m

Angle of horizontal component = 30°

Thus when the suitcase is inclined horizontally at angle of 30° then the work done is :

Work = 196 × 15 × Cos 30°

$= 2940 \times \frac{ \sqrt{3} }{2}$

$= 2940 \times 0.8$

$= 2352joules$

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BY SCIVIBHANSHU

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Answer 2

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