Two constant forces act on an object of mass m = 4.70 kg object moving in the xy

Question

Two constant forces act on an object of mass m = 4.70 kg object moving in the xy plane as shown in the figure below. Force F1 is 25.5 N at 35.0°, and force F2 is 49.5 N at 150°. At time t = 0, the object is at the origin and has velocity (4.30, 2.95) m/s. 150 35.0 (a) Express the two forces in unit-vector notation. F2 (b) Find the total force exerted on the object. (c) Find the object's acceleration m/s2 Now, consider the instant t = 3.00 s. (d) Find the object's velocity. m/s (e) Find its position.

Explanation / Answer

A)

Just use a little trig. here to solve for the components
F1=(F1cos35)i +(F1sin35)j
find the angle of F2 from the negative X-axis
180-150= 30 degrees therefore
F2=(F2cos30)i + (F2sin30)j
***remember in the second quadrant X is negative and Y is positive
Just keep it in mind for the rest of the problem. so therefore,
F2=(-F2cos30)i + (F2sin30)j

B)

Take the vector sum
F1=(F1coos35)i +(F1sin35)j
F2=(-F2cos30)i + (F2sin30)j
-----------------------------------
F(total)=(-21.98)i + (39.38)j find the angle in between the resultant
using arctan(39.38/-21.98) = -60.8 degrees, your angle is 60.8 degrees
** Find the Magnitude F(total)= SQRT(-21.98)^2 + (39.38)^2= 45.1 N

SO F(total) = (F(total)cos60.8)i + (F(total)sin60.8)j

C) use Newton's Law F(total) =ma *** (remember second quadrant)
a= F(total) / mass = 9.6 m/s^2
a= (9.6*cos60.8)i + (9.6*sin60.8)j = (4.7i + 8.4j)

Now, consider the instant t = 3.00 s
D) Use V= V(initial) + at = (4.8i + 2.95j) + (14.1i + 25.2j) = (18.9i + 28.15j) m/s
          

E) Use X = X(initial) + V(inital)*t + 1/2 at^2
          X = 0 + (14.4i + 8.85j) + (21.15i + 37.8j) = (35.55i + 46.65j) m


F) 1/2 m V(final)^2 = 0.5*4.7*(18.9i + 28.15j)^2 = 2701.6 J
    

G) ½mvi2 + F •R
where R is the change in distance from the initial position t=0 to the final position where t = 3 seconds.

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