Answer
This problem involves work, energy conservation, friction, normal force, and kinetic energy principles. a) The work done by the applied force (188 N) is calculated as follows: W = Fd Cosθ, where W = ? (work done), F = 188 N (applied force), d = 48.7 m (distance), and θ = 28.6° (angle). Consequently, W = (188 N)(48.7 m) Cos 28.6° = 8038.46 J. b) The work done against friction can be derived from the equilibrium in the vertical direction: N + FSinθ = weight of the block (mg). Here, the weight is 19.7 kg x 9.81 m/s² = 193.26 N. Solving for the normal force, we find N = 103.26 N. The frictional force is then calculated as f = μN, leading to f = (0.103)(103.26 N) = 10.64 N. The work done by friction is W_f = fd = (10.64 N)(48.7 m) = 517.96 J. c) The change in kinetic energy using conservation of energy principle is ΔKE = W РW_f = 8038.46 J Р517.96 J = 7520.5 J. d) The final speed is determined with ΔKE = 7520.5 J. Using the formula 1/2 m(v_f² Рv_i²), where mass m = 19.7 kg and initial speed v_i = 0 m/s, we find v_f² = (2 x 7520.5 J) / 19.7 kg, resulting in v_f = ,(763.5 m²/s²) = 27.63 m/s.
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