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A 2.0 kg box rests on a plank that is inclined at an angle of 65° above the horizontal. The upper end of the box is attached to...Question
A 2.0 kg box rests on a plank that is inclined at an angle of 65° above the horizontal. The upper end of the box is attached to a spring with a force constant of 17 N/m, as shown in the figure. If the coefficient of static friction between the box and the plank is 0.30, what is the maximum amount the spring can be stretched and the box remain at rest?
Answers
Henry
M*g = 2kg * 9.8N./kg = 19.6 N. = Wt. of
box.
Fp = 19.6*sin65 = 17.76 N. = Force
parallel to the incline.
Fn = 19.6*Cos65 = 8.28 N.=Perpendicular
to the incline = Normal.
Fs = u*Fn = 0.3 * 8.28 = 2.48 N. = Force
of static friction.
Fap-Fp-Fs = M*a
Fap-17.76-2.48 = M*0 = 0
Fap = 20.24 N. = Force applied.
17 * m = 20.24 N.
m = 1.19 m. = Max. distance.
box.
Fp = 19.6*sin65 = 17.76 N. = Force
parallel to the incline.
Fn = 19.6*Cos65 = 8.28 N.=Perpendicular
to the incline = Normal.
Fs = u*Fn = 0.3 * 8.28 = 2.48 N. = Force
of static friction.
Fap-Fp-Fs = M*a
Fap-17.76-2.48 = M*0 = 0
Fap = 20.24 N. = Force applied.
17 * m = 20.24 N.
m = 1.19 m. = Max. distance.