Work and acceleration of a touchdown

  • Thread starter Archetype2
  • Start date
  • #1
Archetype2
4
0

Homework Statement


After scoring a touchdown, an
84.0-kg wide receiver celebrates by leaping
1.20 m off the ground. How much work
was done by the wide receiver in the
celebration?


Homework Equations


W= FD


The Attempt at a Solution


I know that the correct answer should be:
(84.0 kg)(9.80 m/s2)(1.20 m)=988 J
I'm just wondering why the acceleration for the wide-receiver is 9.8m/s^2.
If his acceleration is 9.8 and the acceleration of gravity is 9.8, wouldn't the total acceleration be 0? Also, when a person jumps, he starts off with a positive velocity which decreases, so shouldn't his acceleration be negative?
 

Answers and Replies

  • #2
grzz
998
15
Once he is in the air, the only force acting on him is his weight i.e. the pull of gravity on him. But weight W = mg.

Hence if we use F = ma (downwards) for the man

we get

W = ma
mg = ma
i.e.g=a.
 
  • #3
Archetype2
4
0
OK I get that the acceleration should be 9.8, because there's only one force, but why is it positive instead of negative?
 
Last edited:
  • #4
grzz
998
15
The acceleration due to gravity is always towards the centre of the Earth. In our case we can say that the acc due to gravity is always downwards.

Hence if an object is going upwards this DOWNWARD acceleration due to gravity will decrease this velocity and so it acts as a retardation while if the object is going downwards this DOWNWARDS acceleration due to gravity will increase the velocity.
 

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