A man of mass 50 kg is standing in an elevator- If the elevator is moving up with an acceleration g3 then work done by the normal reaction of the floor of the elevator on a man when the elevator moves by a distance 12 m is- g - 10 m-s2

The correct option is D 8000 J N=m g+m a=m g+m g / 3 N=m(4 g/3)=50 × 4 × 10/3 Now W=F × d=N × d= W=50 × 4 × 10/3× 12=8000 J ...

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Standard VIII

Physics

Free Body Diagram

A man of mass...

Question

A man of mass $50 k g$ is standing in an elevator. If the elevator is moving up with an acceleration $\frac{g}{3}$ then work done by the normal reaction of the floor of the elevator on a man when the elevator moves by a distance $12 m$ is: $(g = 10 m / s^{2})$

2000 J

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4000 J

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6000 J

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8000 J

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Solution

The correct option is D $8000 J$
$N = m g + m a = m g + m g / 3$
$N = m (\frac{4 g}{3}) = \frac{50 \times 4 \times 10}{3}$
Now $W = F \times d = N \times d =$
$W = \frac{50 \times 4 \times 10}{3} \times 12 = 8000 J$

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A man of mass 50 kg is standing in an elevator. If the elevator is moving up with an acceleration g3 then work done by the normal reaction of the floor of the elevator on a man when the elevator moves by a distance 12 m is:(g=10m/s2)

The correct option is D 8000 JN=mg+ma=mg+mg/3N=m(4g3)=50×4×103Now W=F×d=N×d=W=50×4×103×12=8000J

A man of mass $50 k g$ is standing in an elevator. If the elevator is moving up with an acceleration $\frac{g}{3}$ then work done by the normal reaction of the floor of the elevator on a man when the elevator moves by a distance $12 m$ is: $(g = 10 m / s^{2})$

The correct option is D $8000 J$
$N = m g + m a = m g + m g / 3$
$N = m (\frac{4 g}{3}) = \frac{50 \times 4 \times 10}{3}$
Now $W = F \times d = N \times d =$
$W = \frac{50 \times 4 \times 10}{3} \times 12 = 8000 J$