Accelerating Against Friction with Varying
Force

## Question:

A 20 kg block sits on a horizontal surface. The coefficient of
kinetic friction between the block and the surface is 0.3. A
force acts in a constant direction on the block at a downward
angle whose tangent is 3/4 (the force and its vertical and
horizontal components form a 3,4,5 right triangle). The magnitude
of the force increases with distance traveled, s, as F=50*s^{2}.
When s=0, the block has speed 2 m/s. Determine the speed of the
block after it slides 3 meters.
## Answer:

The little bit of work, dWa, done by the applied force in tiny
distance covered, ds, will be the magnitude of the force at that
value of s times the distance ds times the cosine of the angle
between the force and the displacement.
dWa=50*s^{2}*ds*4/5 = 40*s^{2}ds. The work done by the force
of friction during that same travel, ds, is minus the magnitude
of the frictional force times ds. The magnitude of the frictional
force is the coefficient of kinetic friction times the normal
force, Fn, holding the block against the surface. Fn = weight + downward applied force = 20 kg *9.8
m/s^{2} + 50*s^{2}*3/5. Putting the friction force all
together then we get Ff =
0.3*(20*9.8+50*s^{2}*3/5) = 58.8 + 9*s^{2}. The work done by
friction, dWf, (negative work), over distance ds, is dWf = - Ff*ds = - (58.8+9*s^{2})*ds. The total
work, dWt, done over distance ds is dWa+dWf =
(40*s^{2}-(58.8+9*s^{2}))*ds. dWt = (31*s^{2}-58.8)*ds. This is
also the change in kinetic energy over the small distance ds.
The total change in kinetic energy as a function of distance
traveled will be the sum of all the little ds changes. This sum
may be calculated by integrating the expression for dWt with
respect to s from s=0 to s=3. Change in KE =
(10.33*3^3-58.8*3) = 278.9-176.4 = 102.5 Joules. This
positive change in KE means the velocity will increase such that
1/2*m*V2^{2}-1/2*m*V1^{2}=102.5 Joules. This
yields V2^{2}-V1^{2}=10.25 (m/s)^{2}, but V1^{2}=4(m/s^{2}) so
V2^{2}=14.25(m/s)^{2} or V2=3.77
m/s.

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