A block of mass m = 1.0 kg is sent up by the spring which is compressed by x = 1

Question

A block of mass m = 1.0 kg is sent up by the spring which is compressed by x = 10.0 cm. Its path is on the frictionless surface until it reaches the horizontal section CD of length L = 20.0 m with the coefficient of the kinetic friction

A block of mass m = 1.0 kg is sent up by the spring which is compressed by x = 10.0 cm. Its path is on the frictionless surface until it reaches the horizontal section CD of length L = 20.0 m with the coefficient of the kinetic friction mu k = 0.2. The indicated heights are H = 2 m and h = 1.0 m, and both springs have the same spring constant k = 104 N/m. (a) What is the speed of the block at point A? (b) What is the speed of the block at point B? (c) What is the speed of the block at point C? (d) What is the kinetic energy of the block at point D? (e) On its return trip, where between points C and D will the block stop? (f) What should be the compression length of the left spring for a block to slide up just to point B?

Explanation / Answer

a) spring energy = Kinetic energy

0.5*kx^2 = 0.5*m*va^2

va = sqrt(kx^2/m)

va = sqrt(10^4*0.1^2/1) = 10 m/s

b) KEi = KEf + PE

0.5*m*va^2 = mgH + 0.5*m*vb^2

0.5*10^2 = 9.8*2 + 0.5*vb^2

50 - 19.6 = 0.5*vb^2

60.8 = vb^2

vb = 7.79 m/s

c) at point C

KE = KEb + dPE

KE = 0.5*mvb^2 + mg(H-h)

0.5*m*vc^2 = 0.5*1*60.8 + 1*9.8*(2-1)

0.5*vc^2 = 40.2

vc^2 = 80.4

vc = 8.966 m/s

d) at CD friction acts u = 0.2

KEc - work done by friction = KEd

0.5*mvc^2 - u*mg*d = KEd

0.5*1*80.4 - 0.2*1*9.8*20 = KEd

KEd = 1 J

e) Work done = 1J

friction*s = 1J

0.2*1*9.8*s = 1

s = 0.51 m

f) to reach B

0.5*kx^2 = mgH

0.5*10^4*x^2 = 1*9.8*2

x = 0.06261 m

x = 6.261 cm

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