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Set 3 Problem number 12
An object of mass 7.9 kilograms is acted upon by a
net force of 47.4 Newtons.
- The object is initially at rest.
- During the first 3.5 seconds, based on its velocity
change, by how much will its KE increase?
- During the first 3.5 seconds, based on the distance
it travels, how much work is done by the net force on the object?
The acceleration of a 7.9 Kg object under the influence of a 47.4 Newton force will be
a=F/m=( 47.4 Newtons)/( 7.9 Kg)= 6 meters per second per second.
- In 3.5 seconds the velocity will be 3.5( 6 m/s^2) = 21 meters per
second.
- The average velocity will be the average of this velocity and zero, or
( 21 + 0)/2 meters per second = `vAve meters per second.
- At this average velocity, in 3.5 seconds the object will move 36.75
meters.
- The work done will be the product of the distance 36.75 meters and the 47.4
Newton force, or 1741.95 Joules.
- The kinetic energy will be KE = .5( 7.9 kg)(`speed m/s)^2 = 1741.95 Joules.
If an object of mass m, initially at rest, is acted
upon by a net force F for time interval `dt, it will experience acceleration a = F / m for
time interval `dt. This will result in a velocity change `dv = a `dt = F / m * `dt. Since
the object started from rest its final velocity will be
vf = 0 + `dv = F / m * `dt
and its average velocity will be
vAve = (v0 + vf ) / 2 = (0 + F / m * `dt) / 2 =
(1/2) (F / m) `dt.
The distance traveled by the object will be
`ds = vAve `dt = (1/2) (F / m) `dt * `dt = (1/2) (F
/ m) * `dt^2.
The work done on the object will be
`dW = F `ds = F (1/2) (F / m) * `dt^2 = (1/2) F^2 /
m * `dt^2.
The figure below shows the complete relationship
between the work done by the net force and the kinetic energy gained by the object.
- If we find `ds as done in previous problems, we can
then combine this result with the known force F to obtain the work `dW.
- As determined earlier, if v0 = 0 then `ds = (1/2) (F
/ m) * `dt^2.
- After a little substitution and simplification we
will find that `dW = F `ds = (1/2) F^2 / m * `dt^2.
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