A brief note on Work and Kinetic Energy . Read it before a test. Try each question at the end before you open its answer.
The big ideas Scalar product : A ⃗ ⋅ B ⃗ = A B cos θ = A x B x + A y B y + A z B z \vec{A} \cdot \vec{B} = AB\cos\theta = A_xB_x + A_yB_y + A_zB_z A ⋅ B = A B cos θ = A x B x + A y B y + A z B z , a plain number. Work of a steady force: W = F ⃗ ⋅ d ⃗ = F d cos θ W = \vec{F} \cdot \vec{d} = Fd\cos\theta W = F ⋅ d = F d cos θ , in joules. Positive with the motion, zero across it, negative against it. Kinetic energy : K = 1 2 m v 2 = p 2 2 m K = \tfrac12 mv^2 = \dfrac{p^2}{2m} K = 2 1 m v 2 = 2 m p 2 . Work-energy theorem : the net work equals the change in kinetic energy. A changing force : the work is the area under the force-position graph, W = ∫ F d x W = \int F\,dx W = ∫ F d x . A spring: 1 2 k ( x 2 2 − x 1 2 ) \tfrac12 k(x_2^2 - x_1^2) 2 1 k ( x 2 2 − x 1 2 ) . The facts at a glance Joule: 1 J = 1 N m = 1 kg m 2 s − 2 1\ \text{J} = 1\ \text{N m} = 1\ \text{kg m}^2\,\text{s}^{-2} 1 J = 1 N m = 1 kg m 2 s − 2 Work from parts: W = F x d x + F y d y + F z d z W = F_xd_x + F_yd_y + F_zd_z W = F x d x + F y d y + F z d z Braking distance: d = v 2 2 μ g d = \dfrac{v^2}{2\mu g} d = 2 μg v 2 : twice the speed, four times the distance Spring from zero: W = 1 2 k x 2 W = \tfrac12 kx^2 W = 2 1 k x 2 Momentum and K K K : K = p 2 2 m K = \dfrac{p^2}{2m} K = 2 m p 2 A worked example A 1100 kg 1100\ \text{kg} 1100 kg car speeds up from rest to 16 m/s 16\ \text{m/s} 16 m/s over 120 m 120\ \text{m} 120 m of level road.
Kinetic energy gained. 1 2 × 1100 kg × ( 16 m/s ) 2 = 140 800 J \tfrac12 \times 1100\ \text{kg} \times (16\ \text{m/s})^2 = 140\,800\ \text{J} 2 1 × 1100 kg × ( 16 m/s ) 2 = 140 800 J . Net work. By the theorem, 140 800 J 140\,800\ \text{J} 140 800 J . Average net force. 140 800 J 120 m = 1173 N \dfrac{140\,800\ \text{J}}{120\ \text{m}} = 1173\ \text{N} 120 m 140 800 J = 1173 N . Practice with new numbers
A program to try The program works out the work of a slanting pull and the speed it gives.
Predict the work, then run it.
Change the angle to 60 ∘ 60^\circ 6 0 ∘ . By what fraction does the final speed fall?
Mistakes to avoid Work with no displacement along the force. Carrying a bag level, or holding it still, does no work on it. Forgetting the sign. Friction and braking forces do negative work. ( v − u ) 2 (v - u)^2 ( v − u ) 2 for the change in K K K . It is 1 2 m ( v 2 − u 2 ) \tfrac12 m(v^2 - u^2) 2 1 m ( v 2 − u 2 ) : square first. F d Fd F d for a changing force. Use the area under the graph. km/h in 1 2 m v 2 \tfrac12 mv^2 2 1 m v 2 . Change to m/s first. Check yourself A 20 N 20\ \text{N} 20 N pull at 30 ∘ 30^\circ 3 0 ∘ above the floor moves a toy 9.0 m 9.0\ \text{m} 9.0 m . How much work does it do? Stuck? Show a hint Show the answerFind A ⃗ ⋅ B ⃗ \vec{A} \cdot \vec{B} A ⋅ B for A ⃗ = 2 i ^ + 7 j ^ \vec{A} = 2\hat{i} + 7\hat{j} A = 2 i ^ + 7 j ^ and B ⃗ = − 3 i ^ + j ^ \vec{B} = -3\hat{i} + \hat{j} B = − 3 i ^ + j ^ . Stuck? Show a hint Show the answerA 0.50 kg 0.50\ \text{kg} 0.50 kg ball's speed drops from 12 m/s 12\ \text{m/s} 12 m/s to 8.0 m/s 8.0\ \text{m/s} 8.0 m/s . How much work did air resistance do? Stuck? Show a hint Show the answerA spring with k = 600 N/m k = 600\ \text{N/m} k = 600 N/m is stretched from 0 cm 0\ \text{cm} 0 cm to 5.0 cm 5.0\ \text{cm} 5.0 cm . How much work does it take? Stuck? Show a hint Show the answerTwo cars have the same momentum. One has twice the mass of the other. Which has more kinetic energy? Stuck? Show a hint Show the answer