A mass of 9 kg is resting on a horizontal, frictionless surface. Force 1 is appl

Question

A mass of 9 kg is resting on a horizontal, frictionless surface. Force 1 is applied to it at 40 degrees below the horizontal in the -x direction, force 2 has a magnitude of 20 N and is applied in the +x direction , and force 3 has a magnitude of 54 N and is applied vertically upwards. When these forces are applied, the normal force acting on the mass is one half the amount of its weight. Also, when these forces are applied, the object moves 5 meters in the +x direction before it is launched horizontally off a cliff which is 24 meters above the ground. How far does the object travel horizontally in the air, in meters, when it hits the ground? Note: in midair, the only force acting on it is gravity.

Explanation / Answer

when the mass is at height 24 m

along vertical

Fnet = 0

normal force + force3 + Fg   + F1y =0

w/2 j + 54 j - w j - F1*sin40 j = 0

(9*9.8)/2 + 54 = 9*9.8 + F1*sin40

F1 = 15.4 N

force 1 = F1*cos40 i - F1*sin40 j = 15.4*cos40 i - 15.4*sin40 j = 11.8 i- 9.9 j

force 2 = 20 i

force 3 = 54 j

Fnet = force 1 + force 2 + force 3 + normal force + Fg

Fnet = F1x + F1y + F3 + n + Fg + F2

Fnet = F1x + F2 + 0

Fnet = -F1*cos40 + force 2

Fnet = -15.4*cos40 i + 20 i

Fnet = 12 i

acceleration ax = Fnet/m = 12/9 = 1.33 i m/s^2

the mass moves along horizontal direction

during the motion of 5 meters along +x

initial velocity vox = 0

final velocity vx = v

ax = 1.33m/s^2

displacement x = 5 m

vx^2 - vox^2 = 2*ax*x

v^2-0 = 2*1.34*5

v = 3.65 m/s

the mass flies of the cliff with a horizontal speed of 3.65 m/s

after flying off the cliff

along vertical

initial velocity voy = 0

diplacement y = -24 m

acceleration ay = -9.8 m/s^2

y = voy*t + (1/2)*ay*t^2

-24 = 0 -(1/2)*9.8*t^2

t = 2.21 s

along horizontal

accleration ax = 0

x = v*t + (1/2)*ax*t^2

x = 3.65*2.21

x = 8.1 m <<<-------answer

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