A runner whose mass is 49 kg accelerates from a stop to a speed of 6 m/s in 3 seconds. (A good sprinter can run 100 meters in about 10 seconds, with an average speed of 10 m/s.)
(a) What is the average horizontal component of the force that the ground exerts on the runner's shoes?
(b) How much displacement is there of the force that acts on the sole of the runner's shoes, assuming that there is no slipping?
c) How much work is done on the point-particle system by this force?
(a): Force = Mass * Acceleration
Acceleration = Velocity / time
Force = ?
Mass = 49kg
Velocity = 6m/s
Time = 3sec
Distance = 100m
Acceleration = 6/3 = 2m/s^2
Force = (49kg) * (2m/s^2) = 98N
(b)
(c) W = Force * distance
W = Ek
Ek = 1/2 * mass * velocity^2
Ek = 1/2 * 49 * 6 * 6 = 882 Joules
# The Chemical Energy Decreases as the Kinetic Energy of the Runner Increases
(a) To find the average horizontal component of the force, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a).
Given:
Mass (m) = 49 kg
Acceleration (a) = (change in velocity) / (time) = (6 m/s - 0 m/s) / 3 s = 2 m/s^2
Using the formula F = m * a, the force exerted on the runner's shoes can be calculated as follows:
F = 49 kg * 2 m/s^2 = 98 N
Therefore, the average horizontal component of the force is 98 N.
(b) To determine the displacement of the force that acts on the sole of the runner's shoes, assuming no slipping, we need to know the distance traveled by the runner during the acceleration phase.
Given:
Initial speed (u) = 0 m/s
Final speed (v) = 6 m/s
Time (t) = 3 s
Using the formula for displacement (s) in uniformly accelerated motion:
s = (v^2 - u^2) / (2a)
Plugging in the values:
s = (6^2 - 0^2) / (2 * 2) = 18 m
Therefore, the displacement of the force acting on the sole of the runner's shoes is 18 meters.
(c) To find the work done on the point-particle system by the applied force, we can use the work-energy theorem, which states that the work done (W) is equal to the change in kinetic energy (ΔKE).
Given:
Mass (m) = 49 kg
Initial velocity (u) = 0 m/s
Final velocity (v) = 6 m/s
Using the formula for kinetic energy (KE):
KE = (1/2) * m * v^2
Initial kinetic energy (KE1) = (1/2) * 49 kg * 0 m/s = 0 J
Final kinetic energy (KE2) = (1/2) * 49 kg * (6 m/s)^2 = 882 J
ΔKE = KE2 - KE1 = 882 J - 0 J = 882 J
Therefore, the work done on the point-particle system by the force is 882 Joules.
To answer these questions, we can use the principles of force, acceleration, and work.
(a) To find the average horizontal component of the force exerted by the ground on the runner's shoes, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).
Given: mass (m) = 49 kg, final velocity (vf) = 6 m/s, time (t) = 3 seconds, initial velocity (vi) = 0 m/s.
First, we need to calculate the acceleration of the runner using the formula:
acceleration (a) = (change in velocity) / (time)
a = (vf - vi) / t
a = (6 m/s - 0 m/s) / 3 s
a = 2 m/s^2
Now we can find the average horizontal component of the force:
F = ma
F = 49 kg * 2 m/s^2
F = 98 N
Therefore, the average horizontal component of the force that the ground exerts on the runner's shoes is 98 N.
(b) To find the displacement of the force that acts on the sole of the runner's shoes, we can use the work-energy principle, which states that the work done on an object is equal to the force applied on the object multiplied by the distance over which the force is applied (W = F*d).
Given: force (F) = 98 N, displacement (d) = ?
Since there is no slipping, the force exerted on the sole of the runner's shoes is in the horizontal direction. Therefore, the displacement will be along the horizontal direction.
Assuming that the runner starts from rest (initial velocity = 0), we can use the equations of motion to find the displacement:
vf^2 = vi^2 + 2ad
Since the runner starts from rest, the equation simplifies to:
vf^2 = 2ad
Rearranging the equation, we get:
d = vf^2 / (2a)
d = (6 m/s)^2 / (2 * 2 m/s^2)
d = 18 m
Therefore, the displacement of the force that acts on the sole of the runner's shoes is 18 meters.
(c) To find the work done on the point-particle system by the force, we can use the work-energy principle (W = F*d).
Given: force (F) = 98 N, displacement (d) = 18 m
Work (W) = F * d
W = 98 N * 18 m
W = 1764 Joules (J)
Therefore, the work done on the point-particle system by the force is 1764 Joules (J).
force * time = change of momentum
a) force = 49 * (6-0) / 3
= 98 N
b) force acts on the average over one hundred meters
c) work = force * distance = 98 * 100 = 9800 Joules
interesting, lets see how much kinetic energy we have
(1/2) m v^2 = .5 * 49 * 36 = 882 Joules
s we are not very efficient with our fluctuating force.