Question

A ball of mass \(0.2 \mathrm{~kg}\) is thrown vertically upwards by applying a force by hand. if the hand moves \(0.2 \mathrm{~m}\) while applying the force and the ball goes up to \(2 \mathrm{~m}\) height further. find the magnitude of the force. (Consider \(\mathrm{g}=10 \mathrm{~ms}^{-1}\) ) (A) \(16 \mathrm{~N}\) (B) \(20 \mathrm{~N}\) (C) \(22 \mathrm{~N}\) (D) \(4 \mathrm{~N}\)

Short answer

The magnitude of the force applied on the ball is (B) \(20\,\mathrm{N}\).

Step-by-step solution

Calculate the work done by the applied force

First, we need to find the work done by the applied force. Work is calculated as the product of the force, distance, and the cosine of the angle between the force and the distance. Since the force is applied vertically and the distance is also vertical, the angle is 0 degrees, and the cosine of 0 degrees is 1. Therefore, the work done by the applied force can be represented as follows: \(W = F \times d\) where: \(W\) = work done \(F\) = force applied \(d\) = distance force is applied (0.2 m)

Calculate the change in kinetic energy

Since the ball is initially at rest and reaches its maximum height when it has zero velocity, the change in kinetic energy is zero.

Calculate the potential energy gained by the ball

The potential energy gained by the ball as it moves to a height of 2 meters is given by the following formula: \(U = m \times g \times h\) where: \(U\) = potential energy gained \(m\) = mass of the ball (0.2 kg) \(g\) = acceleration due to gravity (10 m/s²) \(h\) = height gained by the ball (2 m) Calculating the potential energy gained: \(U = 0.2 \times 10 \times 2\) \(U = 4\,J\)

Apply the work-energy theorem

According to the work-energy theorem, the work done on an object is equal to its change in kinetic energy plus its gain of potential energy. In this case, the change in kinetic energy is zero. Hence, the work done by the force can be written as: \(W = \Delta K + U\) \(F \times d = 0 + U\)

Calculate the magnitude of the force

Now we can solve for the force by rearranging the last equation: \(F = \frac{U}{d}\) Plugging in the values, we get: \(F = \frac{4\,J}{0.2\,m}\) \(F = 20\,N\) Hence, the magnitude of the force applied is 20 N. The correct answer is (B) \(20\,\mathrm{N}\).

Key concepts

Potential energy

Potential energy is a type of energy stored in an object due to its position or configuration. In this exercise, the ball gains potential energy when it moves upwards against the force of gravity.
To calculate the potential energy (\( U \) ) gained by the ball, use the formula:

• \( U = m \times g \times h \)

Here, \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity, and \( h \) is the height the ball is lifted.
The ball's mass is 0.2 kg, gravity is considered to be 10 m/s², and the ball moves to a height of 2 meters.

• \( U = 0.2 \times 10 \times 2 = 4 \, \text{Joules} \)

This 4 Joules is the potential energy the ball has gained by reaching that height.

Kinetic energy

Kinetic energy is the energy an object possesses due to its motion. In this problem, we analyze the kinetic energy during the ball's journey upwards and when it stops.
Initially, as the ball is thrown, it has some velocity, but when it reaches its maximum height, the velocity becomes zero.

• This is because all the initial kinetic energy has been converted to potential energy at the peak.
• The change in kinetic energy, when the ball begins from rest and stops at the height, is zero since the initial and final velocities are both zero.

Thus, this conversion, or rather the lack of change, shows the energy transfer from kinetic to potential energy.

Vertical motion

Vertical motion refers to the movement of an object up or down in a straight line. The motion of the ball in this problem is purely vertical as it is thrown straight upward by a hand applying force.
This type of motion involves both potential and kinetic energy exchanges.
When analyzing vertical motion:

• Check whether the object is experiencing constant speed or acceleration (due to gravity).
• Both the work-energy principle and equations of motion can describe this movement.

In the ball's case, the force applied gives it initial kinetic energy, and as it moves against gravity, that energy becomes potential energy.

Gravitational force

Gravitational force is the attractive force between two masses. Here, the Earth's gravitational force impacts the ball.
Whenever an object is in vertical motion, gravity acts to pull it downward.

• Gravity's consistent influence means we often consider it constant at 9.8 m/s² or, like in this exercise, approximated as 10 m/s².
• This force creates the conditions for the potential energy as the object gains height.

For the ball, gravitational force is what the applied force must overcome to propel it upwards; thus impacting the calculations of work, force applied, and potential energy gained.