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7.A 1.8 kg block is moved at constant speed over a surface forwhich H 0.25. The displacement is 2 m. It is pulled by aforce directed at 45° to the horizontal as shown in figure. Findthe work done on the block by(a) the force F(b) friction(c) gravity |
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Answer» M = 1.8 kgμ = 0.25 F acts at 45⁰ with horizontal.s = displacement horizontally = 2 meters.Frictional force along the horizontal in the direction opposite to the displacement = Ff = μ * N = μ m g = 0.25 * 1.8 kg * 10 m/s² = 4.5 NewtonsWork done by the Frictional force = Wf = Ff . s = 4.5 * (-2 meters) = - 9 Joules (it is negative as displacement is in the opposite direction to friction). The component of force F, along the horizontal in the direction of s : F Sin 45⁰ = F/√2Since, the block moves with a uniform speed, there is no net force on the block. Hence F/√2 = Ff = 4.5 Newtons F = 4.5 √2 Newtons Work done by F on the block = F . s = F/√2 * 2 = 9 Joules. This is positive as the force does positive work on the block. Gravitational force, ie., the weight and the normal force N are perpendicular to the displacement vector. Hence, the dot product is 0, for the work done. |
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