The figure shows two packages that start sliding down a 20 ramp from rest a dist

Question

The figure shows two packages that start sliding down a 20 ramp from rest a distance d = 6.8 m along the ramp from the bottom. Package A has a mass of 5.0 kg and a coefficient of kinetic friction 0.20 between it and the ramp. Package B has a mass of 10 kg and a coefficient of kinetic friction 0.15 between it and the ramp. How long does it take package A to reach the bottom?

The figure shows two packages that start sliding down a 20½ ramp from rest a distance d = 6.8 m along the ramp from the bottom. Package A has a mass of 5.0 kg and a coefficient of kinetic friction 0.20 between it and the ramp. Package B has a mass of 10 kg and a coefficient of kinetic friction 0.15 between it and the ramp. How long does it take package A to reach the bottom?

Explanation / Answer

For A:
m1g cos(a) = R1 ...(1)
m1g sin(a) + S - u1R1 = m1a1 ...(2)

For B:
m2g cos(a) = R2 ...(3)
m2g sin(a) - S - u1R2 = m2a2 ...(4)

Substitute for R1 from (1) in (2):
m1g [sin(a) - u1 cos(a)] + S = m1a1 ...(5)
g[sin(a) - u1 cos(a)] + [S / m1] = a1 ...(6)

Substitute for R2 from (3) in (4):
m2g [sin(a) - u2 cos(a)] - S = m2a2 ...(7)
g[sin(a) - u2 cos(a)] - [S / m2] = a2 ...(8)

Considering each package alone, and therefore putting S = 0 in (6) and (8), the fact that u1 > u2 means that a1 < a2.
The packages will therefore stay in contact, and we may equate the accelerations putting a2 = a1.

Making this substitution in (7), and adding (5) & (7):
g[ m1(sin(a) - u1 cos(a) ] + m2(sin(a) - u2 cos(a)) ] = (m1 + m2)a1
g[(m1 + m2)sin(a) - (m1u1 + m2u2)cos(a)] = (m1 + m2)a1

a1 = g[(m1 + m2)sin(a) - (m1u1 + m2u2)cos(a)] / (m1 + m2)

Substituting the numbers:
a1 = 9.81 [(5+ 10)sin(20) - (5 * 0.2 + 10* 0.15)cos(20)] / (5 + 10)
= 1.818 m/s^2

The time t to reach the bottom therefore satisfies:
6.8 = a1 t^2 / 2
t^2 = 13.6 / 1.818
t = sqrt(13.6/1.818)
= 2.735 sec

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