Drawing the free body diagram we have d n fs fk m g

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Unformatted text preview: ystem Therefore we can write ΔE = E f − E i = W NC Here the non-conservative force is the frictional force. Drawing € the free-body diagram we have d N € € FS fk m €g The sum of the forces in the y-direction give that € N = mg € From the free-body diagram we can see that the displacement and the frictional force are in opposite directions from each other. So we can write the work done by the non-conservative force as W NC = f k d cos180 = −µk mgd Therefore we can write 0 € 1 212 ΔE = − mv + kd = W NC = −µk mgd 2 2 This is a quadratic equation, solving for d yields € −µk mg ± (µk mg) 2 + kmv 2 d= k = −0.058 m, 0.055 m Since the spring is compressed we take the positive distance. € Extra Example 2: Consider the track below. Section AB is one quadrant of a circle with radius R and is frictionless. BC has a length d and has a coefficient of kinetic friction. CD is frictionless. The block of mass m is released from rest at A. After sliding on the track, the spring is compressed by a distance x. What is the spring constant k? m = 1.0 kg µk = 0.25 d = 3.0 m x = 0.20 m, R = 2.0 m There are two ways to do this problem, the first is to break up the motion of the block into several sections. The second is to do the entire trip in one glorious step. First we will split it up into sections. For section A to B. Here we define the bottom of the ramp as the y=0 position, hence at the bottom of the ramp the block has zero g ravitational potential energy. Since on the ramp there is no frictional force the system in this region is conservative, so we can write ΔE = ΔKE + ΔPE = 0 12 mv 0 − 0 + 0 − mgR = 0 2 Solving for the velocity gives v 0 = 2 gR Section B to C. Here the initial velocity of the block is the velocity € we just found (hence the subscript 0). Since the spring is compressed the block must have a velocity when leaving the regio...
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