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Chapter 3: Problem 328

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

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

Step by step solution

01

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)
02

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.
03

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: \(U = 0.2 \times 10 \times 2\) \(U = 4\,J\)
04

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: \(W = \Delta K + U\) \(F \times d = 0 + U\)
05

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) \(20\,\mathrm{N}\).

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