Question

6. Find out the power of an engine if it lifts a load of 200 kg to a height of 50 m in 10 sec
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Answer 1

Given :

Mass of object = 200kg

Height = 50 m

Time interval = 10 s

To Find :

Power of the engine.

Solution :

❖ Power is defined as the rate of work done per unit time.

Mathematically, P = W / t

• P denotes power
• W denotes work done
• t denotes time

» It is a scalar quantity having only magnitude.

» SI unit : watt (W)

First of all we need to find work done by engine.

It is measured as the product of mass and displacement$.$

» In this case force acting on the object is weight force which is equal to mg.

➠ W = F • d

➠ W = mg • d

➠ W = 200 × 10 × 50

➠ W = 1,00,000 J = 100 kJ

♦ Calculation of power :

➙ P = W / t

➙ P = 100 / 10

➙ P = 10 kW

Answer 2

Answer:

Question

Find out the power of an engine if it lifts a load of 200 kg to a height of 50 m in 10 sec

Given

• mass = 200 kg
• height = 50 metre
• time = 10 sec

To find

power of an engine

Solution

we know that Power is defined as :-

${ \underline{ \boxed{ \sf \large \color{grey}{p = \frac{w}{t}}}}}$

where

• p denotes power
• w denotes work done
• t denotes time

We need to find the work done (w) in order to find the power (p)

work done is given by:-

$\implies{ \underline{ \boxed{ \sf \large \color{grey}{ w = f \times s}}}} \\ \\ \implies{ \sf{ \large{w = m \times g \times \: s }}}$

where

• w denotes work done
• f denotes force
• m denotes the mass
• g denotes the gravity
• s denotes the displacement

Subsitute the value

$\sf{ \to \: w = 20 \times 10 \times 50} \\ \\ \sf{ \to \:w = 100000 \: J = 100 \: kJ}$

Now, we will calculate power :-

$\sf{ \to \: p = \frac{100}{10} } \\ \\ \sf{ \to \:p = 10k W}$

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MoRE To KnOW!!!

☀️Power is defined as rate of work done per unit time

☀️power is a scalar quantity.( it only has magnitude )

☀️power SI unit is watt.

☀️work is defined as product of force and displacement.

☀️ weight = mass × gravity.

☀️work done SI unit is Joule.

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