A block of mass M placed inside a box descends vertically with acceleration ‘a’. The block exerts a force equal to one-fourth of its weight on the floor of the box.

The value of ‘a’ will be

Option 1 -

$\frac{g}{4}$

Option 2 -

$\frac{g}{2}$

Option 3 -

$\frac{3 g}{4}$

Option 4 -

3 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 3

Detailed Solution:

According to Newton’s law of motion, we can write

ma = mg – N = mg - $\frac{m g}{4} = \frac{3 m g}{4}$

$\Rightarrow a = \frac{3 g}{4}$

0 Upvotes 0 Downvotes

Similar Questions for you

A mass of 1 kg is suspended by a thread. It is lifted up with an acceleration of 4.9 ms^–2, and then lowered with an acceleration of 4.9 ms^–2 . The ratio of tensions in first case to second cases is (g = 9.8 ms^–2 )

alok kumar singh

T₁ = m (g + a)

T₂ = m (g - a)

Ball released from height 10 m strikes ground and rebounds height 5 m. Find impulse imparted by ground while collision, given mass of ball is 100 g. (Take g = 10 m/s²)

alok kumar singh

$v_{1} = \sqrt[]{2 g 10}$

$v_{2} = \sqrt[]{2 g 5}$

$\vec{l} = Δ \vec{p}$

$l = 0.1 {\sqrt[]{2 g 10} + \sqrt[]{2 g 5}}$

$= 0.1 {10 \sqrt[]{2} + 10}$

$= (\sqrt[]{2} + 1) N s$

A ball is projected from ground with initial momentum p at an angle q. What is the magnitude of net change in linear momentum when it reaches at maximum height of journey?

alok kumar singh

When an elevator is moving downward with a increasing speed, the apparent weight of a body inside elevator

alok kumar singh

Apparent weight = mg – ma

An insect of mass m = 2 g is inside a vertical drum of radius 1 m that is rotating with an angular velocity of 10 rad/s. If the insect does not fall off then the minimum coefficient of friction is