Two constant forces act on an object of mass m = 5.60 kg object moving in the xy

Question

Two constant forces act on an object of mass m = 5.60 kg object moving in the xy plane as shown in the figure below. Force F1 is 27.5 N at 35.0°, and force F2 is 41.0 N at 150°. At time t = 0, the object is at the origin and has velocity (4.70i + 2.55j) m/s 150° 35.0 (a) Express the two forces in unit-vector notation (b) Find the total force exerted on the object. (c) Find the object's acceleration m/s Now, consider the instant t = 3.00 s (d) Find the object's velocity m/s (e) Find its position (f) Find its kinetic energy from ½mvf2 (g) Find its kinetic energy from ½mvi2 F·

Explanation / Answer

(a)

F1 = 27.5*cos35i + 27.5*sin35 j

F1 = 22.5 i + 15.8 j

F2 = 41*cos150 i + 41*sin150 j = -35.5 i + 20.5 j

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(b)

Fnet = F1 + F2 = -13 i + 36.3 j

magnitude = sqrt(13^2+36.3^2) = 38.6 N

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(c)

acceleration a = Fnet/m = (-13i + 36.3 j)/5.6 = -2.32 i + 6.48 j

magnitude = sqrt(2.32^2+6.48^2) = 6.88 m/s^2

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(d)

v = vi + a*t

v = 4.7i + 2.55j + (-2.32i + 6.48 j)*3

v = 4.7 i + 2.55j - 6.96 i + 19.4 j

v = -2.26i + 21.95 j    m/s

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(e)

r = vi*t + (1/2)*a*t^2

r = (4.7 i + 2.55j)*3 - (1/2)* (-2.32i + 6.48 j)*3^2

r = 4.7i + 2.55 j - 10.44i + 29.16 j

r = -5.74i + 31.71 j

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kinetic energy Kf = (1/2)*m*vi^2 + F.dr

Ki = (1/2)*5.6*(4.7^2+2.55^2) = 80.06 J

F.dr = (-13 i + 36.3 j).( -5.74i + 31.71 j )

F.dr = (13*5.74) + (36.3*31.71) = 1225.7 J

kinetic energy = 80.06 + 1225.7 = 1305.76 J = 1.305 kJ <<<<<<----ANSWER

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