Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.
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Work done</h3>
Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;
W = F × d
Where F is force applied or Weight and d is distance
Also Force = Weight = mass × acceleration due to gravity.
Since gravity is acting on the boxes as it been lift
W = Weight × height from ground level
W = mg × d
Where m is mass of the boxes, g is accelration due to gravity( g = 9.8m/s² ) and d is distance from ground level.
Given the data in the question;
- Since each box has a mass of 7.89 kg
- Mass of the 345 boxes = 345 × 7.89 kg = 2722.05kg
- Distance or height d = 6.0m
To determine the work done, we substitute our values into the expression above.
W = mg × d
W = 2722.05kg × 9.8m/s² × 6.0m
W = 160056.5kgm²/s²
W = 160056.5J
W = 1.6 × 10⁵J
Therefore, Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.
Learn more about work done here: brainly.com/question/26115962
Now I can actually edit my answer directly: I'm fairly sure I've got this wrong, and my mind has gone blank for how to do it, if someone could delete this that would be great and I'll think about it and see if I can figure it out!
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
