Work, Energy and Power

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Work is the scalar product of force and displacement. It is defined as: $$ W = \vec{F} \cdot \vec{S} = FS\cos\theta $$ Where: - $ F $ = magnitude of force (in N) - $ S $ = magnitude of displacement …

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Given: - $ F = 10 \,\text{N} $ - $ S = 5 \,\text{m} $ - $ \theta = 60^\circ $ Formula: $$ W = FS\cos\theta $$ Step 1: $ \cos 60^\circ = 0.5 $ Step 2: $ W = 10 \times 5 \times 0.5 = 25 \,\text{J} $ …

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Work done is zero when: 1. Displacement is zero ($ S = 0 $) 2. Force is perpendicular to displacement ($ \theta = 90^\circ \Rightarrow \cos\theta = 0 $) Examples: - Pushing a wall that doesn't move -…

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Given: - $ m = 4 \,\text{kg} $ - $ h = 3 \,\text{m} $ - $ g = 9.8 \,\text{m/s}^2 $ Force required = weight = $ mg = 4 \times 9.8 = 39.2 \,\text{N} $ Work done = $ F \times S = 39.2 \times 3 = 117.6 …

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The work-energy theorem states that: > The work done by the net force on a body is equal to the change in its kinetic energy. $$ W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 $$ Th…

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Given: - $ m = 5 \,\text{kg} $ - $ u = 6 \,\text{m/s} $ - $ v = 0 \,\text{m/s} $ Initial KE = $ \frac{1}{2}mu^2 = \frac{1}{2} \times 5 \times 6^2 = 90 \,\text{J} $ Final KE = $ \frac{1}{2}mv^2 = 0 $ …

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Kinetic energy (KE) is given by: $$ KE = \frac{1}{2}mv^2 $$ Where: - $ m $ = mass (kg) - $ v $ = speed (m/s) Dependence: - Directly proportional to mass: double mass ⇒ double KE - Proportional to s…

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Given: - $ m = 0.02 \,\text{kg} $ - $ v = 400 \,\text{m/s} $ $$ KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.02 \times (400)^2 $$ Step 1: $ 400^2 = 160000 $ Step 2: $ \frac{1}{2} \times 0.02 = 0.01 $…