1. (a) What horizontal force T:must be applied to pull the supporting rope aside
ID: 776523 • Letter: 1
Question
1. (a) What horizontal force T:must be applied to pull the supporting rope aside until the angle at the wall will be 45 degrees as shown (b) Find T T1 T2 2. A car moving at 30m/s slows uniformly to a speed of 10 m/s in a time of 5s. Determine (a) the acceleration of the car and (b) the distance it moves in 3 seconds and (c) the distance it moves in the third second, 3. A stone is thrown straight upward with a speed of 20m/s. It is caught on its way down at a point 5.0m above where it was thrown. (a) How long did the trip take? (b) How fast was it going when it was caughi? AS -4. An airplane at aheight of 1.000m releases a projectile with an initial velocity of 100m/s at an angle of 37 degrees with the horizontal as shown. (a) Find the time of impact. (b) Find horizontal range. (c) Find v and vy at impact. (d) Find v at impact (magnitude and direction). 100mls 1000mExplanation / Answer
In equilibrium Fnet = 0
along vertical
T1*cos45 = 80
T1 = 80/cos45
alongt horizontal
Fnet = 0
T2 - T1*sin45 = 0
T2 = T1*sin45
T2 = (80/cos45)*sin45 = 80*tan45
T2 = 80 lb
(b)
T1 = 80/cos45 = 113.3 lb
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2)
acceleration a = (vf - vi)/t
acceleration a = (10 - 30)/5 = -20/5 = -4 m/s^2
(b)
distance x = vi*t + (1/2)*a*t^2
x = 30*3 - (1/2)*4*3^2
x1 = 72 m
(c)
distance moved in 2 seconds
x2 = 30*2 - (1/2)*4*2^2 = 52 m
distance moved in 3rd second = x1-x2 = 20 m <<------ANSWER
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3)
initial position y0 = h
final position y = h+5
initial velocity voy = 20 m/s
acceleration ay = -g = -9.8 m/s^2
from equation of motion
y - y0 = v0y*t + (1/2)*ay*t^2
5 = 20*t - (1/2)*9.8*t^2
time t = 3.81 s
(b)
vy^2 - voy^2 = 2*ay*(y-y0)
vy^2 - 20^2 = -2*9.8*5
vy = 17.4 m/s
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4)
along vertical
initial velocity voy = -100*sin37
acceleration ay = -g = -9.8 m/s^2
initial position yo = 1000 m
final position y = 0
from equation of motion
y - yo = voy*t + (1/2)*ay*t^2
-1000 = -100*sin37*t - (1/2)*9.8*t^2
t = 9.41 s
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(b)
along horizontal
initial velocity vox = 100*cos37
acceleration ax = 0
x = vox*t + (1/2)*ax*t^2
x = 100*cos37*9.41
x = 751.5 m
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(c)
vx = vox + ax*t
vx = 100*cos37 = 79.8 m/s
vy = voy*ay*t
vy = -100*sin37 - 9.8*9.41 = -152.4 m/s
-------------------------
(d)
v = sqrt(vx^2+vy^2)
v = 172 m/s
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