Lifting a 50-kg sack through a vertical distance of 2 m requires same amount of work as lifting a 25-kg sack through a vertical distance of 4 m
<h3>How to determine which will require more work</h3>
To determine which will require more work, we shall determine the work done in lifting each sack. Details below:
i. Work done in lifting 50 Kg
- Mass (m) = 50 Kg
- Height (h) = 2 m
- Acceleration due to gravity (g) = 9.8 m/s²
- Workdone (Wd) =?
Wd = mgh
Wd = 50 × 9.8 × 2
Workdone = 980 J
ii. Work done in lifting 25 Kg
- Mass (m) = 25 Kg
- Height (h) = 4 m
- Acceleration due to gravity (g) = 9.8 m/s²
- Workdone (Wd) =?
Wd = mgh
Wd = 25 × 9.8 × 4
Workdone = 980 J
SUMMARY
- Work done in lifting 50 Kg is 980 J
- Work done in lifting 25 Kg is 980 J
Thus, equal amount of work done is needed.
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Fam I believe the answer..... is it states that for every particle found there is an identical "mirror" particle that is the same in every way except for spin
Answer:
Fscos63
Explanation:
Given that a horizontal pole is attached to the side of a building. There is a pivot P at the wall and a chain is connected from the end of the pole to a point higher up the wall. There is a tension force F in the chain. What is the moment of the force F about the pivot P?
Taking the moment from the pivot point P, that means the moment at point p = 0
Then, if we consider the weight mg of the pole, according to the principle of equilibrium : sum of the upward forces equal to the sum of the downward forces.
Therefore, mg = Fsinø ....... (1)
Also, taking moment at point P
Let the length of the pole = s
The length of the weight of the pole = 1/2 S
Fscosø = mgs/2
The distance s will cancel out
2Fcosø = mg ...... (3)
Substitute mg in equation 1 into equation 3
2fcosø = fsinø
F will cancel out
Tanø = 2
Ø = tan^-1(2)
Ø = 63.4 degree
The moment of force F about pivot point P will be
Moment = force × distance
Moment = Fcos63 × S
Moment = Fscos63