WebSep 12, 2024 · Figure : Work done by a constant force. (a) A person pushes a lawn mower with a constant force. The component of the force parallel to the displacement is the … WebFigure 7.3 Work done by a constant force. (a) A person pushes a lawn mower with a constant force. The component of the force parallel to the displacement is the work done, as shown in the equation in the figure. (b) A person holds a briefcase. No work is done because the displacement is zero.
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Web(a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the lift by the gravitational force in this process? (c) What is the total work done on the lift? 26. WebAs expected, the net work is the net force times distance. Solution for (a) The net force is the push force minus friction, or F net = 120 N – 5. 00 N = 115 N. Thus the net work is W net = F net d = 115 N 0.800 m = 92.0 N ⋅ m = 92.0 J. 7.16 Discussion for (a) This value is the net work done on the package. dunn purnsley actor
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WebMar 16, 2024 · We define it as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains its kinetic energy unless its speed … WebThis is a statement of the work–energy theorem, which is expressed mathematically as. W = Δ K E = 1 2 m v 2 2 − 1 2 m v 1 2. The subscripts 2 and 1 indicate the final and initial velocity, respectively. This theorem was proposed and successfully tested by James Joule, shown … WebWork done, transfer of energy, work done formula, examples and step by step solutions, GCSE / IGCSE Physics, notes. Work Done by a Force. ... A car drives 10 m up a hill at constant speed, as shown, If its weight is 15 kN and there is a frictional force of 4 kN acting between its tyres and the road surface, calculate: ... dunnow hall