Two constant forces act on an object of mass m = 4.90 kg object moving in the xy

Question

Two constant forces act on an object of mass m = 4.90 kg object moving in the xy plane as shown in the figure below. Force F rightarrow 1 is 29.0 N at 35.0degree, and force F rightarrow 2 is 48.5 N at 150 degree. At time t = 0, the object is at the origin and has velocity (3.70i^ + 2.30j^ m/s. Express the two forces in unit-vector notation. Find the total force exerted on the object. Find the object's acceleration. Now, consider the instant t = 3.00 s. Find the object's velocity. Find its position Find its kinetic energy from 1/2mvf2 Find its kinetic energy from 1/2mvi2 + ?F rightarrow · ?rvec

Explanation / Answer

A)

F1=29cos(35) i +29sin(35) j= 23.7554i +16.634j

F2=48.5cos(150) i +48.5sin(150) j=-42i+24.25j

B)F=F1+F2=-18.245i+40.884j

|F|=44.77N

C)a=F/m=-3.723i +8.344j

So

|a|=9.14m/s^2

D)v=u+at

So at t=3

v=3.7i+2.3j -11.169i+25.032j =-7.47i +27.332j

So

|v|=28.33m/s

E)x=ut+0.5at^2

So

x=11.1i+6.9j -16.7545i+37.55j

=-5.6545i +44.45j

F)0.5mvf^2=1966 J (approx)

G)0.5mvi^2+F.r=46.501+(18.245*5.6545+40.884*44.45)=1966J (approx)

Answers to F and G are equal as expected

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