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.
Important Questions from Work, Energy, and EMF
Which of the following is NOT a true difference between EMF and potential difference (PD)?
A battery of EMF 6.0 V and internal resistance 1.0 Ω is connected to a resistor of 11 Ω. The terminal potential difference for the battery is:
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:
Two batteries, E1 (emf: 3 V, internal resistance: 0.5 Ω) and E2(emf: 6 V, internal resistance: 1.0 Ω), are connected in series by connecting the positive terminal of E2 to the negative terminal of E1 . A third battery E3 (emf: 6 V, internal resistance: 1.0 Ω) is connected in parallel with this combination by connecting its positive terminal to the positive terminal of E1 and its negative terminal to the E2. The equivalent emf of this combination is:
- 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
