BOX ON INCLINE HITS SPRING AND REBOUNDS!!!!!!

Question

An 84kg block with an initial velocity of 3.6 m/sec slides 228cm down an incline of 18.4 degrees and makes contact with a spring, compressing the string 14cm, then rebounds up the incline a distance of 136cm beyond the position of the relaxed spring and comes to a stop.

a) Determine the coefficient of friction between the incline and the sliding block.

b)Then determine the spring constant.

Answer 1

This problem is all about keeping track of the energy:

Energy is conserved except for the work of friction on the block.

Let's define zero potential energy as the height where the block just touches the spring.

Initial conditions:

KE = 1/2mv^2 = 544.3 J

PE = mgh = 84*9.81*2.28*sin(18.4) = 593.0 J

Total energy = KE + PE = 1147.3 J

Final conditions:

KE = 0

PE = 84*9.81*1.36*sin(18.4) = 353.7 J

Total energy = 353.7 J

Loss of energy = 1147.3 - 353.7 = 793.6 J

This must be equal to the work done by the friction force on the sliding block.

Work = force * distance

Distance = 2.28 + 0.14 + 0.14 + 1.36 = 3.92 m

force = 793.6 / 3.92 = 202.4 N

The normal force is equal to the weight of the block * cos(18.4)

Normal force = 84*9.81*cos(18.4) = 781.9 N

Coefficient of friction = 202.4 / 781.9 = 0.26

Energy at max spring compression = Initial energy - fricition work done on the way down.

Energy = 1147.3 - 202.4 * (2.28 + 0.14) = 657.5 J

This will be equal to the potential energy of the block + the potential energy of the spring (no kinetic energy, everything's at rest)

PE block = -84*9.81*0.14*sin(18.4) = -36.4 J (notice the minus sign)

PE spring = 657.5 + 36.4 = 693.9 J

= 1/2kx^2

so k = 70800 N/m