Question

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Answer 1

Force exerted by sprinter = 69.68 N

Step-by-step explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

Answer 2

69.68 N

Step-by-step explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki =
`(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)`

W =
`F_(total) .d`

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki =
`(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)`

`F_(total) .d =(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)`

`F_(total) = ((1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i))/(d)`

`F_(total=) ((1)/(2) X 62 X6^(2) -(1)/(2) X 62 X2^(2) )/(25)`

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

`F_(sprinter) = F_(total) + F_(wind) = 39.7 + 30 = 69.68 N`