Question

Knowing that a 0.90-kg object weighs 9.0 n , find the acceleration of a 0.90-kg stone in free fall .

Answer 1

The acceleration in a free fall would be
`10\; {\rm m\cdot s^(-2)}` at this location.

Step-by-step explanation:

This problem can be solved using Newton's Laws of Motions. By Newton's Laws of Motion, the acceleration of an object can be found by dividing the net force on the object by the mass of the object:

`\displaystyle (\text{acceleration}) = \frac{(\text{net force})}{(\text{mass})}`.

In a free fall, the net force on the object would be equal to gravitational attraction, since that would be the only force acting on the object. Note that gravitational attraction is also knowns as the weight of the object.

`(\text{net force in free fall}) = (\text{weight})`.

The acceleration of the object in the free fall would be:

`\begin{aligned}(\text{acceleration}) &= \frac{(\text{net force})}{(\text{mass})} \\ &= \frac{(\text{weight})}{(\text{mass})}\end{aligned}`.

In other words, the acceleration of the object in the free fall would be equal to the ratio between the weight and mass of the object. For the
`0.90\; {\rm kg}` object with a weight of
`9.0\; {\rm N}`, the free-fall acceleration would be:

`\begin{aligned}(\text{acceleration}) &= \frac{(\text{weight})}{(\text{mass})} \\ &= \frac{9.0\; {\rm N}}{0.90\; {\rm kg}} \\ &= 10\; {\rm m\cdot s^(-2)}\end{aligned}`.

(Note that in standard units,
`1\; {\rm N} = 1\; {\rm kg\cdot m\cdot s^(-2)}`.)

Additionally, at the same location of one particular gravitational field, the acceleration in a free fall is the same regardless of mass. Since the
`0.90\; {\rm kg}` object would accelerate at
`10\; {\rm m\cdot s^(-2)}` in a free fall, the
`0.90\; {\rm kg}` stone would accelerate at the same rate at this location in the gravitational field.