Question

${\Large{\rm{\underline{\underline{\red{Question :}}}}}}$
A man pulls a roller by a force of 20 kg-f applied at 60° with the ground. If he pulls it a distance of 10 m in 1 minute, calculate the power dissipated.
(Take g = 10 m/s²)


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Answer 1

Given :

• Force acted on the roller = 20 kg-f
• Angle formed by force on the ground = 60°
• Distance travelled in 1 min = 10 m
• g = 10 m/s²

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To find :

• The power dissipated.

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Formula to be used :

$\bf\dag$ $\bf\:Work~=~Fd$

where, F is the force and d is the displacement.

$\bf\dag$ $\bf\:Power~developed~in~time~“ t ”~=~\frac{W}{t}$

where, W denotes work done and t denotes time.

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Solution :

We know,

$\bf\:Work~=~Fd$

$\sf\longrightarrow\:Work~=~m~g~Cos(60)~d$

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$\sf\because\:F=m \times g \times Cosθ$

Since force is resolved in two components :

• one perpendicular to incline = Cosθ
• other parallel to incline = Sinθ

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Now substituting the values in -

$\sf\longrightarrow\:Work~=~m~g~Cos(60)~d$

$\sf\longrightarrow\:Work~=~20 \times 10 \times \frac{1}{2} \times 10$

$\sf\longrightarrow\:Work~=~200 \times \frac{1}{2} \times 10$

$\sf\longrightarrow\:Work~=~\frac{200}{2} \times 10$

$\sf\longrightarrow\:Work~=~\cancel{\frac{200}{2}} \times 10$

$\sf\longrightarrow\:Work~=~100 \times 10$

$\sf\longrightarrow\:Work~=~1000~J$

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Finally calculating power dissipated -

$\bf\:Power~developed~in~time~“ t ”~=~\frac{W}{t}$

Substituting the values

$\sf\longrightarrow\:Power~dissipated~=~\frac{1000~J}{1~min}$

$\sf\longrightarrow\:Power~dissipated~=~1000~J/min$

$\small{\underline{\boxed{\mathrm{\longrightarrow~Power~dissipated~=~1000~Watt}}}}$ $\sf\color{cyan}{⋆}$

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Answer 2

Answer:

1kgwt=9.8N

So, force F=20kgwt=20×9.8N=196N

The horizontal component of the force F is Fh

=Fcos60=196cos60=98N

The required work done W=Fh

d=98×10=980J

The power developed in t=1 minute =W/t=980J/60s=16.33 Watt