Question

A 54 kg sprinter, starting from rest, runs 46 m in 7.0 s at constant acceleration.

What is the magnitude of the horizontal force acting on the sprinter?

What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s?

Answer 1

Solution :-

For the force, F = ma

To find the acceleration, apply d = v_ot + .5at²

46 = 0 + .5(a)(7²)

a = 1.877 m/s²

Then, F = ma = (54)(1.877) = 101.358 N

Power is found by P = Fv

We need the velocity at 2, 4, and 6 sec

Apply v_f = v_o + at

v_f = 0 + (1.877)(2) = 3.754 m/s

v_f = 0 + (1.755)(4) = 7.508 m/s

v_f = 0 + (1.755)(6) = 11.262 m/s

So, at 2 sec, P = (101.358)(3.754) = 380.498 W, which is .380 kW

at 4 sec, P = (101.358)(7.508) = 761 W, which is .761 kW

at 6 sec, P = (101.358)(11.262) = 1141.49 W, which is 1.1414 kW