Accelerating Against Friction with Varying Force


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*s2. When s=0, the block has speed 2 m/s. Determine the speed of the block after it slides 3 meters.


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*s2*ds*4/5 = 40*s2ds. 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/s2 + 50*s2*3/5. Putting the friction force all together then we get Ff = 0.3*(20*9.8+50*s2*3/5) = 58.8 + 9*s2. The work done by friction, dWf, (negative work), over distance ds, is dWf = - Ff*ds = - (58.8+9*s2)*ds. The total work, dWt, done over distance ds is dWa+dWf = (40*s2-(58.8+9*s2))*ds. dWt = (31*s2-58.8)*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*V22-1/2*m*V12=102.5 Joules. This yields V22-V12=10.25 (m/s)2, but V12=4(m/s2) so V22=14.25(m/s)2 or V2=3.77 m/s.

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