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MaRussiya [10]
3 years ago
6

A man holds a child on his shoulders while watching a football match. How many work has he done if the game lasts for 2 hours? ​

Physics
2 answers:
Sedaia [141]3 years ago
8 0
Answer: ZERO!

work= F•d = F*d*cos(theta)

Since there is no displacement taking place then there is NO WORK being done! Cheers!
Shalnov [3]3 years ago
4 0

very interesting and amazing question u had asked but i get it...

Answer

  • work done is given by

W = Force × Displacement

  • but in above case force is exerted on the men as the weight of the child × gravitational pull of earth (9.8 m/s^2) but the displacement or movement is ZERO
  • there for W= F×0 =0
  • so even after carrying child for 2 hours work done by the men will be zero

<u>More</u><u> </u><u>examples</u>

  • a person standing straight and lifting weight
  • a person carry some weight and walk in a circular path and came back to its original position.........in this case also Displacement will be zero

------------------------------------

hope it helped u buddy

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ball dropped from top of 50m high cliff stone thrown straight up from bottom with speed of 24. they collide. how far above base
Tresset [83]

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28.73 m from the base of the cliff collide happen

Explanation:

Equation for ball dropped from 50 cliff is reaches distance x in t seconds isi given by

x=ut+\frac{1}{2} at^2\\\\x= 0 \times t+\frac{1}{2} gt^2 =\frac{1}{2} gt^2......(1)

stone is thrown up from the bottom with speed u=24 m/s . it reaches distance y when stone collide with ball.(g is negative here)

y=24t-\frac{1}{2} gt^2.............(2)

we know that total distance traveled by ball and stone is 50 m

x+y=50 m

adding equation 1 and 2, we get time t

x+y=\frac{1}{2} gt^2+24t-\frac{1}{2}gt^2\\50=24t\\t=2.083 s

substitute this time in equation 2, we can get the required distance where they collide

y=24t-\frac{1}{2} gt^2\\y=24\times 2.083-\frac{1}{2}\times  9.8\times 2.083^2\\y=28.73 m

28.73 m from the base of the cliff collide happen

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