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Answer:
The sled took 9.96 seconds to descend the slope.
Explanation:
Refer to the provided figure for a description of the scenario. In red, you can see the x and y-components of the gravitational force (Fg). The y-component of Fg (Fgy) matches the magnitude of Fn but acts in the opposite direction, resulting in the cancellation of both forces. The equations regarding the sled's acceleration arise from the force of the wind (Fw), the friction force (Ff), and the x-component of the gravitational force (Fgx).The totality of these forces causes the sled's motion. Finding the resultant force enables us to ascertain the sled's acceleration, which we can then use to compute the sled's travel time.
The magnitude of the friction force is calculated as follows:
Ff = μ · Fn where:
[[TAG_27]]μ = coefficient of kinetic friction[[TAG_28]][[TAG_29]]Fn = normal force[[TAG_30]][[TAG_31]][[TAG_32]][[TAG_33]]The normal force equals the y-component of the gravitational force:[[TAG_34]][[TAG_35]]Fgy = Fg · cos 30º = m · g · cos 30º[[TAG_36]][[TAG_37]]Where [[TAG_38]][[TAG_39]]m = mass [[TAG_40]][[TAG_41]]g = acceleration due to gravity[[TAG_42]][[TAG_43]]Then:[[TAG_44]][[TAG_45]]Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º[[TAG_46]][[TAG_47]]Fgy = 744 N[[TAG_48]][[TAG_49]][[TAG_50]][[TAG_51]][[TAG_52]]Consequently, the magnitude of Fn is also 744 N, leading to the friction force: [[TAG_53]][[TAG_54]]Ff = μ · Fn = 0.151 · 744 N = 112 N[[TAG_55]][[TAG_56]][[TAG_57]][[TAG_58]]Now, for the x-component of Fg, Fgx is found as follows:[[TAG_59]][[TAG_60]]Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N[[TAG_61]][[TAG_62]][[TAG_63]][[TAG_64]]The resultant force, Fr, is the sum of all these forces:[[TAG_65]]Fw + Fgx - Ff = Fr [[TAG_66]][[TAG_67]](Keep in mind that forces are vectors and the friction's direction opposes the others, thus it must have the opposite sign).[[TAG_68]][[TAG_69]]Fr = 161 N + 430 N - 112 N = 479 N[[TAG_70]][[TAG_71]]With this resultant force calculated, we deduce the sled's acceleration:[[TAG_72]][[TAG_73]][[TAG_74]][[TAG_75]]F = m·a [[TAG_76]][[TAG_77]]where:[[TAG_78]][[TAG_79]]F = force[[TAG_80]][[TAG_81]]m = mass of the object[[TAG_82]][[TAG_83]]a = acceleration[[TAG_84]][[TAG_85]]So:[[TAG_86]][[TAG_87]]F/m = a[[TAG_88]][[TAG_89]]a = 479N/87.7 kg = 5.46 m/s²[[TAG_90]][[TAG_91]][[TAG_92]][[TAG_93]]The equation for the positioning of an object with uniform acceleration on a straight path is as follows:[[TAG_94]][[TAG_95]]x = x0 + v0 · t + 1/2 · a · t²[[TAG_96]][[TAG_97]]where:[[TAG_98]][[TAG_99]]x = position at time t[[TAG_100]][[TAG_101]]x0 = initial position[[TAG_102]][[TAG_103]]v0 = initial velocity[[TAG_104]][[TAG_105]]t = time[[TAG_106]][[TAG_107]]a = acceleration[[TAG_108]][[TAG_109]][[TAG_110]][[TAG_111]]As the sled begins from rest with the reference point where it begins to slide, x0 and v0 = 0.[[TAG_112]][[TAG_113]][[TAG_114]][[TAG_115]]Thus, x = 1/2· a ·t²[[TAG_116]][[TAG_117]][[TAG_118]][[TAG_119]]Let’s determine the time required for the sled to reach 271 m:[[TAG_120]][[TAG_121]]271 m = 1/2 · 5.46 m/s² · t²[[TAG_122]][[TAG_123]]2 · 271 m / 5.46 m/s² = t²[[TAG_124]][[TAG_125]][[TAG_126]]t = 9.96 s [[TAG_127]][[TAG_128]][[TAG_129]]Effectively, it takes nearly 10 seconds for the sled to travel down the slope.[[TAG_130]][[TAG_131]] [[TAG_132]][[TAG_133]][[TAG_134]][[TAG_135]][[TAG_136]][[TAG_137]][[TAG_138]]
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