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Question

What is the work required to increase the speed of a 0.4 kg ball from 1 m/s to 3 m/s?

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
1.6 J

Calculating Work Done Using Work-Energy Theorem

The work required to change an object's speed is equal to the change in its kinetic energy. This is known as the Work-Energy Theorem.

The formula for kinetic energy (KE) is given by:

\(KE = \frac{1}{2}mv^2\)

Where:

  • \(m\) is the mass of the object.
  • \(v\) is the speed of the object.

The work done (\(W\)) is the difference between the final kinetic energy (\(KE_f\)) and the initial kinetic energy (\(KE_i\)):

\(W = \Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\)

Applying the Work-Energy Theorem

Given values:

  • Mass (\(m\)): 0.4 kg
  • Initial speed (\(v_i\)): 1 m/s
  • Final speed (\(v_f\)): 3 m/s

Step 1: Calculate the Initial Kinetic Energy (\(KE_i\))

\(KE_i = \frac{1}{2} \times 0.4 \text{ kg} \times (1 \text{ m/s})^2\)

\(KE_i = \frac{1}{2} \times 0.4 \times 1\)

\(KE_i = 0.2 \text{ J}\)

Step 2: Calculate the Final Kinetic Energy (\(KE_f\))

\(KE_f = \frac{1}{2} \times 0.4 \text{ kg} \times (3 \text{ m/s})^2\)

\(KE_f = \frac{1}{2} \times 0.4 \times 9\)

\(KE_f = 0.2 \times 9\)

\(KE_f = 1.8 \text{ J}\)

Step 3: Calculate the Work Done (\(W\))

\(W = KE_f - KE_i\)

\(W = 1.8 \text{ J} - 0.2 \text{ J}\)

\(W = 1.6 \text{ J}\)

Conclusion

The work required to increase the speed of the ball is 1.6 Joules.

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Important Questions from Work, Energy, and EMF

  1. Which of the following is NOT a true difference between EMF and potential difference (PD)?

  2. Consider two cells of emf ε1 and ε2 with internal resistances r1 and r2, respectively. The two cells are connected in parallel by connecting their positive terminals together and connecting their negative terminals together. The combination is equivalent to a single cell with emf given by:

  3. Two batteries E 1 (emf: 6 V, internal resistance: 0.5 Ω ) and E 2  (emf: 12 V, internal resistance: 1.0  Ω ) are  connected in parallel by connecting their positive terminals to point A and negative terminals to point B. A  third battery E 3  [emf: 6 V, internal resistance: ( \(\frac{2}{3}\) )  Ω ] is connected in series with this combination by  connecting its positive terminal to B. The equivalent emf of this combination is
  4. A particle is moved from point \(X\) to point \(Y\) under a force field. The work done depends only on the initial and final speeds. Which one of the following is correct?
  5. The cost of energy to operate an industrial refrigerator that consumes 5 kW power working 10 hours per day for 30 days will be (Given that the charge k.W.h of energy = Rs. 4)

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