Question

A 6kg object undergoes an acceleration of 2m/s, what is the magnitude of the resultant acting on it . If this same force is applied to a 4kg object, what acceleration is produced

Answer 1

`12\; \rm N`.

`3\; \rm m\cdot s^(-2)`.

Step-by-step explanation:

By Newton's Second Law, the acceleration of an object is proportional to the size of the resultant force on it, and inversely proportional to the mass of this object.

`\displaystyle \text{acceleration} = \frac{\text{resultant force}}{\text{mass}}`.

Rearrange this equation for the resultant force on the object:

`\text{resultant force} = \text{acceleration} \cdot \text{mass}`.

For the
`6\; \rm kg` object in this question:

`\begin{aligned} F &= m \cdot a \\ &= 6\; \rm kg * 2\; \rm m \cdot s^(-2) \\ &=12\; \rm N\end{aligned}`.

When the resultant force on the
`4\; \rm kg` object is also
`12\; \rm N`, the acceleration of that object would be:

`\begin{aligned} a &= (F)/(m) \\ &= (12\; \rm N)/(4\; \rm kg) = 3\; \rm m \cdot s^(-2)\end{aligned}`.