Two constant forces act on a 5.20 kg object moving in the xy plane as shown in t

Question

Two constant forces act on a 5.20 kg object moving in the xy plane as shown in the figure below. Force F_1 is 28.5 N at 35.0 degree, while F_2 is 45.5 N at 150 degree. At time t = 0, the object is at the origin and has velocity (4.60i + 2.65j) m/s. (a) Express the two forces in unit-vector notation. Find the total force on the object. Find the object's acceleration. Now, consider the instant t = 3.00 s. Find the object's velocity. Find the object's location. Find the object's kinetic energy from Find the object's kinetic energy from What conclusion can you draw by comparing the answers to parts and (g)?

Explanation / Answer

mass m = 5.2 kg

F1 = 28.5 cos 35 i + 28.5 sin 35 j

= 23.345 i + 16.346 j

F2 = 45.5 cos 150 i + 45.5 sin 150 j

= -39.4 i + 22.75 j

(b). Total force F = F1+F2

= ( 23.345 i + 16.346 j)+(-39.4 i + 22.75 j )

= -16.055 i + 39.096 j

(c). Accleration a = F / m

= ( -16.055 i + 39.096 j )/5.2

= -3.0875 i + 7.518 j

(d).Initial velocity u = 4.6 i + 2.65 j

Object velocity at t=3 s is v = u + at

v = (4.6 i + 2.65 j )+(-3.0875 i + 7.518 j)(3)

= (4.6 i + 2.65 j ) +(-9.2625 i +22.555 j )

= -4.6625 i + 25.205 j

(e). object location S = ut +(1/2) at 2

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