Work done = force * distance
work done = 200 * 20
work done = 4000J
Explanation:
Below is an attachment containing the solution.
The relationship between the masses of the Earth, moon and sun and their distances to each other play critical roles in affecting tides
- Mass of the elevator (m) = 570 Kg
- Acceleration = 1.5 m/s^2
- Distance (s) = 13 m
- Let the force be F.
- We know, F = ma,
- Therefore, F = (570 × 1.5) N = 855 N
- Angle between distance and force (θ) = 0°
- We know, work done = F s Cos θ
- Therefore, work done by the cable during this part
- = (855 × 13 × Cos 0°) J
- = (855 × 13 × 1) J
- = 11115 J
<u>Answer</u><u>:</u>
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Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
Explanation:
Sam mass=75kg
Height is 50m
20° frictionless slope
Horizontal force on Sam is 200N
According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.
Therefore
Wg - Ww =∆K.E
Note initial the body was at rest at top of the slope.
Then, ∆K.E is K.E(final) - K.E(initial)
K.E Is given as ½mv²
Since initial velocity is zero then, K.E(initial ) is zero
Therefore, ∆K.E=½mVf²
Wg is work done by gravity and it is given by using P.E formulas
Wg=mgh
Wg=75×9.8×50
Wg=36750J
Ww is work done by wind and it's is given by using formulae for work
Work=force × distance
Ww=horizontal force × horizontal distance
Using Trig.
TanX=opposite/adjacent
Tan20=h/x
x=h/tan20
x=50/tan20
x=137.37m
Then,
Ww=F×x
Ww=200×137.37
We=27474J
Now applying the formula
Wg - Ww =∆K.E
36750 - 27474 =½×75×Vf²
9276=37.5Vf²
Vf²=9275/37.5
Vf²= 247.36
Vf=√247.36
Vf=15.73m/s
