MECHANICAL WORK, ENERGY, AND POWER - Biomechanics and Motor Control of Human Movement - page 147

Since both force and velocity are vectors, we must take the dot product,
or the product of the force and the component of the velocity that is in the
same direction as the force. This will yield:

P = FV cos θ = F x V x + F y V y (6.6)

where: θ = angle between force and velocity vectors in the plane

defined by those vectors

F x and F y = forces in x and y directions

V x and V y = velocities in x and y directions

For the purpose of this initial discussion, let us assume that the force and the
velocity are always in the same direction. Therefore, cos θ = 1 and:

P = FV W

t t

W = Pdt = FV dt J (6.7)

0 0

Example 6.1. A baseball is thrown with a constant accelerating force of
100 N for a period of 180 ms. The mass of the baseball is 1.0 kg, and it starts
from rest. Calculate the work done on the baseball during the time of force
application.

Solution

1

2 at 2

S 1 = ut +

u = 0

100 m/s 2 a = F /m = 100 / 1 . 0 =

1

100 ( 0 . 18 ) 2 S 1 =

2 ×

= 1 . 62 m

S 1

W = Fds = FS 1 = 100 × 1 . 62 = 162 J

0

Example 6.2. A baseball of mass 1 kg is thrown with a force that varies with
time, as indicated in Figure 6.7. The velocity of the baseball in the direction
of the force is also plotted on the same time base and was calculated from the
time integral of the acceleration curve (which has the same numerical value
as the force curve because the mass of the baseball is 1 kg). Calculate the
instantaneous power to the baseball and the total work done on the baseball
during the throwing period.
The peak power calculated here may be considered quite high, but it should
be noted that this peak has a short duration. The average power for the
throwing period is less than 500 W. In real-life situations, it is highly unlikely
that the force will ever be constant; thus, instantaneous power must always be