The kinetic friction coefficient is listed on the left side of the table. Thanks

Question

The kinetic friction coefficient is listed on the left side of the table. Thanks for flagging it.

A light spring of force constant 3.70 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.430 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is the following. In each case, assume that the coefficient of static friction is greater than the coefficient of kinetic friction. Let the positive direction point to the right. 0.000 m/s m/s 0.100 m/s m/s 0.445 m/s m/s

Explanation / Answer

The force exerted by the spring on each block is in magnitude

Fs = kx = 3.7 * 8 x 10^-2 m = 0.296 N

part a )

With no friction, the elastic energy in the spring becomes kinetic energy of the blocks, which have momenta of equal magnitude in opposite directions. The blocks move with constant speed after they leave the spring. From conservation of energy,

(K+U)i =(K +U)f

kx^2/2 = m1v1f^2/2 + m2v2f^2/2

1/2 * (3.7)*(0.08)^2 = 1/2 * (0.250)v1f^2 + 1/2 * (0.430)*v2f^2 ...(1)

onservation of linear momentum

m1v1i + m2v2i = m1v1f+m2v2f

0 = (0.250)v1f (-i) + (0.430)v2f(i)

v1f = 1.72v2f

substitue this value into eq (1)

0.01184 = 1/2 * (0.250)(1.72v2f)^2 + 1/2 * (0.430)v2f^2

0.01184 = 0.5848 v2f^2

v2f = 0.14 m/s

v1f = 1.72 * v2f = 0.2408 m/s

0.250 kg = 0.2408 m/s

0.430 kg = 0.14 m/s

part b )

uk = 0.100

for lighter block

forces in y direction = m*ay

n - 0.250 * 9.8 = 0

n = 2.45 N

fk = uk*n = 0.1*2.45 = 0.245 N

the maximum force of static friction is a similar size. Since 0.296 N is larger than 0.245 N, this block moves

for the heavier block the normal force and the frictional force are = uk*n = 0.1 * 0.430 * 9.8 = 0.4214 N since 0.296 N is less than this force so heavier block won't move .

In this case, the frictional forces exerted by the floor change the momentum of the two-block system. The lighter block will gain speed as long as the spring force is larger than the friction force: that is until the spring compression
becomes xf given

Fs = kx = 0.245 N = 3.7 * xf

xf = 0.0662 m

the energy of the lighter block as it moves to this maximum-speed point

Ki + Ui - fk*d = Kf + Uf

0 + 0.01184 - 0.245 * (0.08 - 0.0662) = 1/2 * 0.250 * vf^2 + 1/2 * 3.7 * 0.0662^2

8.459 x 10^-3 = 1/2 * 0.250 * vf^2 + 8.107514 x 10^-3

vf = 0.053 m/s

for 0.250 kg = 0.053 m/s

for 0.430 kg = 0 m/s

part c )

uk = 0.445

for the lighter block fk = 0.445 * 2.45 = 1.09025 N

The force of static friction must be at least as large. the 0.296 N spring force is too small to produce motion of either block

0.250 kg = 0m/s

0.430 kg = 0 m/s

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