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Question

What is the force required to produce an acceleration of 9.8 m/s 2on a body of weight 9.8N? Take g = 9.8 m/s 2.

The correct answer is

9.8 N

This question asks us to find the force needed to give a specific acceleration to a body of a given weight. To solve this, we need to use the concepts of weight, mass, and Newton's Second Law of Motion.

Understanding Weight and Mass

Weight is the force of gravity acting on a body's mass. It is calculated using the formula:

\(W = mg\)

Where:

  • \(W\) is the weight of the body (in Newtons, N)
  • \(m\) is the mass of the body (in kilograms, kg)
  • \(g\) is the acceleration due to gravity (in meters per second squared, m/s²)

Mass, on the other hand, is a measure of the amount of matter in a body and is independent of gravity. We can find the mass of the body from its given weight and the value of \(g\).

Calculating the Mass of the Body

Given:

  • Weight \(W = 9.8 \text{ N}\)
  • Acceleration due to gravity \(g = 9.8 \text{ m/s}^2\)

Using the formula \(W = mg\), we can rearrange it to find the mass \(m\):

\(m = \frac{W}{g}\)

Plugging in the given values:

\(m = \frac{9.8 \text{ N}}{9.8 \text{ m/s}^2}\)

\(m = 1 \text{ kg}\)

So, the mass of the body is 1 kg.

Applying Newton's Second Law of Motion

Newton's Second Law of Motion states that the force \(F\) required to accelerate a body is directly proportional to its mass \(m\) and the acceleration \(a\) produced. The formula is:

\(F = ma\)

Where:

  • \(F\) is the net force applied (in Newtons, N)
  • \(m\) is the mass of the body (in kilograms, kg)
  • \(a\) is the acceleration produced (in meters per second squared, m/s²)

Calculating the Required Force

We need to find the force required to produce an acceleration of 9.8 m/s² on the body. We know the mass of the body and the desired acceleration:

  • Mass \(m = 1 \text{ kg}\) (calculated above)
  • Required acceleration \(a = 9.8 \text{ m/s}^2\)

Using Newton's Second Law \(F = ma\):

\(F = (1 \text{ kg}) \times (9.8 \text{ m/s}^2)\)

\(F = 9.8 \text{ N}\)

Therefore, the force required to produce an acceleration of 9.8 m/s² on a body of weight 9.8 N is 9.8 N.

Given Information Value
Weight (\(W\)) 9.8 N
Acceleration due to gravity (\(g\)) 9.8 m/s²
Required acceleration (\(a\)) 9.8 m/s²

Step Calculation Result
1. Find Mass (\(m\)) \(m = W/g = 9.8 \text{ N} / 9.8 \text{ m/s}^2\) 1 kg
2. Find Force (\(F\)) \(F = ma = 1 \text{ kg} \times 9.8 \text{ m/s}^2\) 9.8 N

Final Answer Summary

The force required to produce the specified acceleration is 9.8 N.

Revision Table: Force, Mass, Weight, Acceleration

Let's quickly review the key concepts and formulas used in this problem:

Concept Definition Formula
Weight (\(W\)) Force of gravity on a mass \(W = mg\)
Mass (\(m\)) Amount of matter in a body \(m = W/g\) (derived)
Force (\(F\)) Push or pull that can cause acceleration \(F = ma\) (Newton's 2nd Law)
Acceleration (\(a\)) Rate of change of velocity \(a = F/m\) (derived)

Additional Information: Units and Laws

Understanding the units is crucial in physics calculations. The standard unit for force is the Newton (N), for mass is the kilogram (kg), and for acceleration is meters per second squared (m/s²). Newton's Second Law (\(F = ma\)) is one of the fundamental laws of classical mechanics and forms the basis for understanding how forces affect the motion of objects. This problem is a direct application of this law combined with the definition of weight.

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Important Questions from Acceleration due to gravity of the earth

  1. If a ball is thrown vertically upwards with the speed $u$, the distance covered during the second-to-last $t$ seconds of its ascent is
    (Assume $t$ is less than half of the total ascent time).
  2. A body freely falling from rest has acquired a velocity ‘v’ after it falls through a distance ‘h’. The distance it has to fall down further for its velocity to become double is:

  3. On earth, the value of G = 6.67 × 10 -11  Nm 2kg -2 . What is the value on moon, where acceleration due to gravity is nearly one - sixth than that of earth?

  4. At what height above the surface of the earth does the weight of an object reduce by 1%. Given the radius of the earth is 6400.

  5. The radii of two planets are respectively R 1and R 2and their densities are respectively ρ 1and ρ 2. The ratio of the acceleration due to gravity (g 1/g 2) at their surface is

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