18. The following force vectors act on an object: 25 Newtons at 45. north of eas

Question

18. The following force vectors act on an object: 25 Newtons at 45. north of east and ii) SO Newtons at 30, south of east. Which of the following represents the magnitude of the resultant and its angle relative to the casterly direction? A) 61.4 Newtons -6.85 deg B) 55.6 Newtons-7.50 deg C) 61.4 Newtons 21.8 deg D) 35.7 Newtons O deg E) 75.0 Newtons 15 deg 19. A runner of mass 56.1 kg starts from rest and accelerates with a constant acceleration of I m/s until she reaches a velocity of 4.3 m/s. She then continues running a this constant velocity. How long in seconds does the runner take to travel 120 m? A) 26.4 sec B) 30.06 sec C) 35.8 sec D) 12.9 sec E) 13.333333 sec 20. A bridge that was 8 m long has been washed out by the rain several days ago. How fst must a car be going to succesfuly jump the strceanm" Alhough the roxd is level o both sides of the bridge, the road on the far side is 2.5 m lower than the road on this side. A) 11.2 m/s B) 76.8 m/s C) 25.6 m/s D) 16 m/s E) 6.25 m/s

Explanation / Answer

18) A) 61.4 N, -6.85 deg

Let East be +x axis

Fnetx = 25*cos(45) + 50*cos(30)

= 60.97 N

Fnety = 25*sin(45) - 50*sin(30)

= -7.32 N

Fnet = sqrt(Fnetx^2 + Fnety^2)

= sqrt(60.97^2 + 7.32^2)

= 61.4 N

theta = tan^-1(Fnety/Fnetx)

= tan^-1(-7.32/60.97)

= -6.85 degrees

19) B) 30.06 s

time taken to get 4.3 m/s velocity, t = (4.3 - 0)/1 = 4.3 s

distance travelled during first 4.3 s, d1 = u*t + (1/2)*a*t^2

= 0 + (1/2)*1*4.3^2

= 9.245 m

remaining distance, d2 = 120 - 9.245

= 110.755 m

remaining time taken = 110.755/4.3

= 25.757 s

total time taken = 4.3 + 25.757

= 30.06 s

20) A) 11.2 m/s

let t is the time taken to land.

use, y = (1/2)*g*t^2

==> t = sqrt(2*y/g)

= sqrt(2*2.5/9.8)

= 0.71428 s

now use, vox = x/t

= 8/0.71428

= 11.2 m/s

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