Calculating power give mass, time, distance

  • Thread starter Thread starter Verloren
  • Start date Start date
  • Tags Tags
    Mass Power Time
AI Thread Summary
To calculate the power produced by a sprinter weighing 56 kg during the first second of a 50-meter race completed in 7 seconds, the final velocity after one second can be determined as 50/7 m/s. The initial velocity is 0, and the work done can be calculated using the equation for kinetic energy, w = (1/2)mvf² - (1/2)mvo². By substituting the values into the power equation P = w/t, where t is 1 second, the power output during the acceleration phase can be determined. Understanding the relationship between distance, velocity, and time is crucial for solving the problem effectively. This approach will yield the required power calculation for the sprinter's performance.
Verloren
Messages
1
Reaction score
0

Homework Statement


So I have a scenario given by my physics teacher that says a sprinter that weighs 56 kg ran a 50 meter race in 7 seconds. For the first second he accelerates, but for the 6 remaining seconds he goes at a constant pace. I'm supposed to calculate the power produced within the first second.

The Attempt at a Solution


I've tried using df=(vo+vf)/2 * t (where df=final distance, vo=initial velocity (0), and vf=final velocity) but since I'm not given the final distance traveled in the one second or the final velocity I'm not sure what to do.

The goal of finding final velocity is to maybe use it in the equation w=(1/2mvf2)-(1/2mvo2) (where m=mass vf=final velocity and vo=initlal velocity) to find work and then plug work into the equation P=w/t (p = power, w=work, t=time).

Any help concerning equations to look at or a relationship that I'm missing would be much appreciated
 
Physics news on Phys.org
Verloren said:
I've tried using df=(vo+vf)/2 * t (where df=final distance, vo=initial velocity (0), and vf=final velocity)d

Hi ! Veloren.

As i see if not going wrong.

The velocity just after one second and after one second is (50/7) m/s.:smile:Let say, the sprinter travel 50 meter in 7s for all distances meanwhile the end of the speed is (50/7) m/s.Therefore, just before the constant speed is equal to final speed instant 1s => Vf= 50/7 m/s.
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top