Asked by ashyLerry
The 235-kg cart in the figure starts from rest and rolls with negligible friction. It is pulled by three ropes as shown. It moves 600. m horizontally. Find the final velocity of the cart.
F1 = 500N at 0 degree
F2 = 500N at 40.1 degree
F3 = 400N at 30 degree
F1 = 500N at 0 degree
F2 = 500N at 40.1 degree
F3 = 400N at 30 degree
Answers
Answered by
Henry
X = hor. = 500 + 500cos40.1 + 400cos30,
X = 500 + 382.5 + 346.4 = 1229N.
Y = ver. = 500sin40.1 + 400sin30,
Y = 322.1 + 200 = 522.1N.
F^2 = X^2 + Y^2,
F^2 = (1229)^2 + (522)^2,
F^2 = 1,783,029.
F = 1335N. = Resultant force.
a = F / m = 1335 / 235 = 5.69m/s^2.
(Vf)^2 = Vo^2 + 2ad,
(Vf)^2 = 0 + 2*5.69*600 = 6828,
Vf = 82.6m/s.
X = 500 + 382.5 + 346.4 = 1229N.
Y = ver. = 500sin40.1 + 400sin30,
Y = 322.1 + 200 = 522.1N.
F^2 = X^2 + Y^2,
F^2 = (1229)^2 + (522)^2,
F^2 = 1,783,029.
F = 1335N. = Resultant force.
a = F / m = 1335 / 235 = 5.69m/s^2.
(Vf)^2 = Vo^2 + 2ad,
(Vf)^2 = 0 + 2*5.69*600 = 6828,
Vf = 82.6m/s.
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