Gravitational Force Calculation
The question asks us to find the approximate value of 'g' (acceleration due to gravity) given the mass of an object and the gravitational force it experiences.
Given Information
- Mass of the object, $m = 5 \, \text{kg}$
- Gravitational force, $F = 49 \, \text{N}$
Relevant Formula
The relationship between force (F), mass (m), and acceleration due to gravity (g) is given by Newton's second law, applied to gravity:
$ F = m \times g $
Solving for g
To find the value of 'g', we can rearrange the formula:
$ g = \frac{F}{m} $
Calculation
Substitute the given values into the rearranged formula:
$ g = \frac{49 \, \text{N}}{5 \, \text{kg}} $
$ g = 9.8 \, \text{N/kg} $
Since $1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2$, the unit N/kg is equivalent to m/s². Therefore:
$ g = 9.8 \, \text{m/s}^2 $
Conclusion
The calculated value of 'g' is approximately $9.8 \, \text{m/s}^2$. This matches option 2.
Important Questions from Universal law of gravitation
Which of the following laws says that "Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them?"
The force of attraction between two objects of masses 'M' and 'm' which lie at a distance 'd' from each other is directly proportional to the-
The force of attraction (F) between two particles having masses m 1and m 2is given by _______. (If r is the distance between them and G is a universal constant)
Three point masses each of mass m are placed at the three corners of an equilateral triangle of side x. Find the resultant force acting on any one particle at the corner.
Imagine a light planet is revolving around a star in a circular orbit of radius R with the period of revolution T . If the gravitational force of attraction between the two is proportional to R(-5/2) then
