You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Physics

Free Body Diagram

The normal re...

Question

The normal reaction on a body placed in a lift moving up with constant acceleration $2 m / s^{2}$ is $120 N$ . Mass of body is (take $g = 10 m / s^{2}$ )

10 k g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

15 k g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12 k g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 k g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $10 k g$
From force diagram shown in figure
$N - m g = m a_{0}$
$120 - m g = m (2)$
$m = \frac{120}{10 + 2} = 10 k g$

Suggest Corrections

Similar questions

Q. A body of mass

10 k g

is placed in a lift moving upward with an acceleration of

2 m / s^{2}

. The apparent weight of body is

(g = 9.8 m / s^{2})

Q. A body of mass

10 k g

is hung by a spring-balance in a lift. What woul be the reading of the balance when (i) the lift is ascending with an acceleration of

2 m / s^{2}

, (ii) descending with the same acceleration, (iii)descending with a constant velocity of

2 m / s

(g = 10 m / s^{2})

Q. Find the force

F

acting on the body of mass

6 kg

to move it with an acceleration of

1.5 {m/s}^{2}

. Take

g = 10 {m/s}^{2}

A body of mass 10kg is hanging from another body of mass 15kg. The combination is pulled up by a string with an acceleration of . $1.2 m^{- 2}$ Find the tensions (in N) at A and B points as shown in figure $(T a k e g = 9.8 m / s^{2})$ .

Join BYJU'S Learning Program

The normal reaction on a body placed in a lift moving up with constant acceleration 2m/s2 is 120N. Mass of body is (take g=10m/s2 )

The correct option is A 10kgFrom force diagram shown in figureN−mg=ma0120−mg=m(2)m=12010+2=10kg

The normal reaction on a body placed in a lift moving up with constant acceleration $2 m / s^{2}$ is $120 N$ . Mass of body is (take $g = 10 m / s^{2}$ )

The correct option is A $10 k g$
From force diagram shown in figure
$N - m g = m a_{0}$
$120 - m g = m (2)$
$m = \frac{120}{10 + 2} = 10 k g$