the mass of a person is 65kg how fast should he run so that the kinetic energy is 812.5j
Answers
Answered by
1
k = 1/2mv²
812.5×2/65 = v²
105625 = v
v = 325 m/s
812.5×2/65 = v²
105625 = v
v = 325 m/s
Answered by
0
b) use the following equation to calculate the minimum speed the high jumper must reach take-off in order to jump over the bar.
Kinetic energy = 1/2 X mass X speed^2
Please help
…see more
Think about how gravitational energy is converted to kinetic energy as an object accelerates.
This is the reverse. The jumper converts kinetic energy to GPE as he gains height.
GPE is at a maximum when the jumpers' kinetic energy is zero at the apex of the jump. i.e. all of his KE is converted to GPE.
To solve: rewrite the equation making V the subject and substituting KE = GPE.
Last edited by uberteknik; 22-04-2014 at 19:43.
1
reply

Electricity
Offline
9
Rep:
follow
3
22-04-2014 20:01
(Original post by uberteknik)
It is correct. GPE = mgh
Think about how gravitational energy is converted to kinetic energy as an object accelerates.
This is the reverse. The jumper converts kinetic energy to GPE as he gains height.
GPE is at a maximum when the jumpers' kinetic energy is zero at the apex of the jump. i.e. all of his KE is converted to GPE.
To solve: rewrite the equation making V the subject and substituting KE = GPE.
…see more
So by apex of the jump you mean the climax of it? So when he jumps and he stops for a bit before declining? All the KE is transferred to GPE, because he stops moving for a moment?
TO make V the subject I think its
V = kinetic energy / mass / 2 then square root?
Do I divide the mass by 2 prior the equation or after the ke/m part?
How do I substitute Ke = GPE I dont understand that part?
Last edited by Electricity; 22-04-2014 at 20:02.
0
reply

uberteknik
Offline
21
Rep:
follow
4
22-04-2014 20:28
(Original post by Electricity)
So by apex of the jump you mean the climax of it?
…see more
Climax is not a term used in physics! But yes, when he reaches the highest point in the jump.
(Original post by Electricity)
So when he jumps and he stops for a bit before declining? All the KE is transferred to GPE, because he stops moving for a moment?
…see more
If you mean GPE is a maximum WHEN he stops moving, then yes.
GPE is the potential energy an object has by means of its height above ground. A person can sit on a window ledge all day long and his gravitational potential energy will be the same all day long because he is not moving.
It is only when he jumps off the ledge that his potential energy starts converting to kinetic energy by means of that gravitational acceleration.
When he hits the ground, GPE is at a minimum (mgh = 0 because h = 0), and his KE is at a maximum (1/2mv2 = max when v = max)
(Original post by Electricity)
TO make V the subject I think its
V = kinetic energy / mass / 2 then square root?
Do I divide the mass by 2 prior the equation or after the ke/m part?
How do I substitute Ke = GPE I dont understand that part?
…see more
You can do it in one of two ways but they are both the same thing:
By making:
mgh = 1/2mv2
Spoiler:
show
Or similarly, 1/2mv2 = 812.2J
Spoiler:
show
Last edited by uberteknik; 22-04-2014 at 20:38.
0
reply

Electricity
Offline
9
Rep:
follow
5
22-04-2014 20:54
(Original post by uberteknik)
Climax is not a term used in physics! But yes, when he reaches the highest point in the jump.
If you mean GPE is a maximum WHEN he stops moving, then yes.
GPE is the potential energy an object has by means of its height above ground. A person can sit on a window ledge all day long and his gravitational potential energy will be the same all day long because he is not moving.
It is only when he jumps off the ledge that his potential energy starts converting to kinetic energy by means of that gravitational acceleration.
When he hits the ground, GPE is at a minimum (mgh = 0 because h = 0), and his KE is at a maximum (1/2mv2 = max when v = max)
You can do it in one of two ways but they are both the same thing:
By making:
mgh = 1/2mv2
Spoiler:
show
Or similarly, 1/2mv2 = 812.2J
Spoiler:
show
…see more
Oh right OK. I get it now thanks!
Similar questions