Question

two horizontal forces of 12n and 8n act on a body of mass 4kg. the minimum value of acceleration produced cannot be less than​

Answer 1

Two horizontal forces of 12N and 8N act on a body of mass 4kg.

The minimum net force experienced by body will be when :

• The forces are directed opposite to one another.

$F_{net} = 12 - 8 = 4 \: N$

$\implies a_{net} = \dfrac{ 4 }{mass}$

$\implies a_{net} = \dfrac{4}{4}$

$\implies a_{net} = 1 \: m {s}^{ - 2}$

So, minimum acceleration of the body is 1 m/s².

Answer 2

Given :  two horizontal forces of 12N  and 8N act on a body of mass 4kg.

To Find : the minimum value of acceleration produced cannot be less than​

Solution:

Forces = 12 N  and  8 N

Forces are Horizontal Hence angle between forces will  be

either 0° or 180°

Means Forces will be in the same direction or opposite direction

If Forces are in the same direction

Then Net forces = 12 + 8 = 20 N

If Forces are in the opposite direction

Then Net forces = 12 - 8 = 4 N

Force = mass * acceleration

mass  = 4 kg

Hence acceleration can be  

20/4 =  5 m/s²

or

4/4 = 1 m/s²

Hence minimum value of acceleration produced cannot be less than​ 1 m/s²

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