Answer:
180 W
Explanation:
The work done by the man against gravity is equal to its gain in gravitational potential energy:
where
(mg) = 720 N is the weight of the man
is the change in height
Substituting,
The power he must deliver is given by
where
W = 3600 J
t = 20 s is the time taken
Substituting,
Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time,
(a) Acceleration,
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :
s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration,
s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
There are 3 forces acting on the stoplight:
• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward
• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right
The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that
∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0
We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to
2<em>T</em> sin(<em>θ</em>) = <em>W</em>
2 (1000 N) sin(<em>θ</em>) = 100 N
sin(<em>θ</em>) = 0.05
<em>θ</em> ≈ 2.87°
If <em>y</em> is the vertical distance between the stoplight and the ground, then
tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)
Solve for <em>y</em> :
tan(2.87°) = (15 m - <em>y</em>) / (100 m)
<em>y</em> = 15 m - (100 m) tan(2.87°)
<em>y</em> ≈ 9.99 m
