Question

22. A porter lifts a luggage of 15 kg from the
ground and put it on the head 1.7 m above the
ground. Find the work done by the porter on the
luggage (g = 10 meter /sec square).
1 Point
*
255J
355J
15)
none of these​

Answer 1

Given :-

Mass of the luggage = 15 kg

Height of the luggage = 1.7 m

To Find :-

The work done by the porter on the  luggage.

Analysis :-

Here we are given with the mass and height of the luggage.

In order to find the work done substitute the values given in the question such that work done is equal to mass into gravity into height and find the work done by the porter on the luggage accordingly.

Solution :-

We know that,

• g = Gravity
• w = Work
• m = Mass
• h = Height

Using the formula,

$\underline{\boxed{\sf Work \ done=Mass \times Gravity \times Height}}$

Given that,

Mass (m) = 15 kg

Gravity (g) = 10 m/s

Height (h) = 1.7 m

Substituting their values,

⇒ w = mgh

⇒ w = 15 × 10 × 1.7

⇒ w = 255 J

Therefore, the work done by the porter on the luggage is 255 J.

Answer 2

Answer:

Given :-

• Mass of luggage = 15 kg
• Height from ground when it put = 1.7 m
• Acceleration due to gravity = 10 m/s²

To Find :-

Work done

Solution :-

$\small {\fbox {\fbox {firstly \: lets \: see \: the \: concept}}}$

Here, is a luggage which is lifting by a porter and the mass of luggage is 15 kg and it is put from the ground about 1.7 m and the acceleration due to gravity is 10 mps and we have to find to the work done by the luggage.

$\small {\fbox {\fbox {lets \: see \: how \: to \: solve}}}$

Here, we will find the work done by a formula which mass×gravity×Height. And when we multiply them then we will got our required answer.

$\huge {\fbox {\fbox {lets \: calculate}}}$

${\boxed {\tt Work \: done = Mass \times Gravity \times Height}}$

Work done = 15 × 1.7 × 10

Work done = 150 × 1.7

Work done = 255 J

Hence :-

Work done by luggage is 255 Joule.

Extra bytes

Joule :-

Joule is the SI unit of energy. The work required to produce one watt of power for one second is also written as Joule.