A 65 kg engineering student is training for a boxing match W

A 65 kg engineering student is training for a boxing match (Welterweight class), and wonders how much power is required during rope skipping. He skips rope 180 times per minute, and 2/5 of the time he is in contact with the ground. Assuming that the ground reaction force is constant when he is on the ground, what is the power during contact with the ground? Assume no energy lost in friction or impact

ime on in ground ai 2/5 time 3/5 time

Solution

He skips 180 times / minute

So, in time of each skip, t = 1/180 = 5.56*10^-3 min = 5.56*10^-3*60 = 0.334 s

So, time in which he is off the ground, t\' = (3/5)*0.334 = 0.2 s

So, So, time taken by the student to reach the top of the its flight,t* = 0.2/2 = 0.1 s

So, t* = 0.1s

Now, initial velocity, u = gt = 9.8*0.1 = 0.98 m/s

So, Kinetic energy of the student at the time of impact,

KE = 0.5*65*u^2 = 0.5*65*(0.98)^2 = 31.2 J

Now, assuming constant force,

Power during contact, P = work done/time

Now, work done = KE at impact <------- By work energy principle

So, P = 31.2/((2/5)*0.334)

So, P = 233.5 W <--------answer