When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 n backward on the ground for 0.800 s. what is his f
1 answer:
You might be interested in
Ok, we need to find a relation for the speed as it relates to the acceleration. This is given by the integral of acceleration:
Where we have the initial velocity is 0m/s and a will be 4.90m/s².
But we see there is an issue now... We know the velocity as a function of time, but we don't know how long the car has been accelerating! We need to calculate this time by now finding the position function as a function of time. This way we can solve for the time, t, that it takes to go 200m accelerating this way and then substitute that time into our velocity equation and get the velocity.
Position is just the integral of velocity:
Where the initial velocity and initial position are both zero.
Now we set this position function equal to 200m and find the time, t, it took to get there
Now let's put t=9.04s into our velocity equation:
Where we have the initial velocity is 0m/s and a will be 4.90m/s².
But we see there is an issue now... We know the velocity as a function of time, but we don't know how long the car has been accelerating! We need to calculate this time by now finding the position function as a function of time. This way we can solve for the time, t, that it takes to go 200m accelerating this way and then substitute that time into our velocity equation and get the velocity.
Position is just the integral of velocity:
Where the initial velocity and initial position are both zero.
Now we set this position function equal to 200m and find the time, t, it took to get there
Now let's put t=9.04s into our velocity equation:
Answer
given,
k = 250 N/m
q = 900 N/m³
(FSp)s=−kΔs−q(Δs)^3
work done = Force x displacement
limits are x = 0 to x = 0.15 m
work done
W = 3.375 + 0.1139
W = 3.3488 J
b) % cubic term =
% cubic term =