Respuesta :
Answer:
69.68 N
Explanation:
Work done is equal to change in kinetic energy
W = ΔK = Kf - Ki = [tex]\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}[/tex]
W = [tex]F_{total} .d[/tex]
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 = [tex]\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}[/tex]
[tex]F_{total} .d =\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}[/tex]
[tex]F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}[/tex]
[tex]F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}[/tex]
= 39.7
we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N
[tex]F_{sprinter} = F_{total} + F_{wind} = 39.7 + 30 = 69.68 N[/tex]
Answer:
Force exerted by sprinter = 69.68 N
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
