HCI 05 WEP Solutions (Lecture Examples)

Solutions to WEP LN Examples

Example 1 A lawn mower is pushed by a force P of 20.0 N at 30 to the horizontal, against a horizontal resistive force D of 8.0 N. It moves a horizontal distance of 10.0 m. Calculate (i) the work done by the force P on the lawn mower. (ii) the work done by the resistive force D. (iii) the work done by the normal contact force from the ground and the weight of the lawn mower. (iv) the change in kinetic energy of the mower. (i) W = Fs cosθ = (20.0)(10.0cos30°) = 173 J (ii) W = Fs cosθ = (8.0)(10.0cos180°) = -80.0 J (iii) W = Fscosθ = (N)(10.0cos90°) = 0 J Since weight is also normal to displacement, Work done by weight = 0 J (iv) Net WD = Total work done by all forces = 173 + (-80.0) + (0) + (0) = 93 J Increase in KE = Net WD = 93 J D

Example 2 A mass of weight 5.0 N is lifted by a vertical force of 7.0 N through a vertical distance of 2.0 m. (a) Calculate the work done by the lifting force. (b) Calculate the work done by the gravitational force. (c) Calculate the change in gravitational potential energy. (d) Calculate the total work done on the mass. (e) Assuming it was initially at rest. calculate the final speed of the mass. W(5.0 N) F (7.0 N) h (2.0 m) (a) Work done by vertical liftng force = F s cosθ = (7.0)(2.0)(1) = +14 J (b) Work done by gravitational force = mg s cosθ = (5.0)(2.0)(-1) = -10 J (c) ΔG.P .E. = mgΔh = 5.0 x 2.0 = +10 J (Gain in GPE) (d) Total work done on mass = Sum of work done by lifting force & weight = 14 + (-10) = 4 J (e) KE gained by mass = total work done on mass = 4 J     22 2 -1 1 ( - ) 42 1 5.0 42 9.81 4 m s m v 0 v v

Example 3 (J99/I/23 mod) A sample is placed in a tensile testing machine. It is extended by known amounts and the tension is measured. (i) What is the work done in stretching the sample to the first point where it no longer obeys Hooke’s Law? (ii) What is the work done on the sample when it is given a total extension of 9.0 mm? = 0.15 + 0.28 = 0.43 J W = Area under the force-displacement graph  60+80+ 0.009 - 0.005 2 (ii) 0.15 W   11 (60)(0.005) = 0.15 J 22 (i) =W Fx

Example 4 (a) A motorcyclist drives horizontally off a cliff at a speed of 38.0 m s-1. Ignoring air resistance, find the speed of the motorcycle just before it reaches the ground. Applying Principle of Conservation of Energy on the motorcyclist, KEi + GPEi = KEf + GPEf 𝑣 = 2𝑔Δℎ + 𝑢2    2 -1 2 9.81 35 38.0 46.2 m s v   GPEi – GPEf = KEf - KEi (Loss in GPE = Gain in KE) mgΔh = ½ m(v2 - u2)

Example 4 b) Would the answer change if the motorcyclist leaps across at the same 35 m height with an initial upward angle while maintaining the same initial speed of 38.0 m s-1? As air resistance is negligible, as long as the motorcycle has fallen by a vertical height of 35 m (regardless of the actual path), the loss in gravitational PE and therefore the gain in KE remains the same.

Example 5 A car of mass 800 kg moving at 30 km h-1 along a horizontal road is brought to rest by