A body of mass 5,kg is suspended by a spring balance on an inclined plane as shown in figure. (in N

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4-1.Newton's Laws of Motion

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A body of mass $5\,kg$ is suspended by a spring balance on an inclined plane as shown in figure. (in $N$)

A$50$

B$25$

C$500$

D$10$

(AIIMS-2018)

Solution

So, force applied on spring balance is
Acceleration of the body down the rough inclined plane $=g \sin \theta$
$\therefore$ Force applied on spring balance
$=m g \sin \theta=5 \times 10 \times \sin 30^{\circ}$
$=5 \times 10 \times \frac{1}{2}=25 N$

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Solution

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