Two packages at UPS start sliding down the 20 degree ramp shown in the figure. P

Question

Two packages at UPS start sliding down the 20 degree ramp shown in the figure. Package A has a mass of 5.50kg and a coefficient of kinetic friction of 0.210. Package B has a mass of 11.5kg and a coefficient of kinetic friction of 0.160.

How long does it take package A to reach the bottom?

Two packages at UPS start sliding down the 20 degree ramp shown in the figure. Package A has a mass of 5.50kg and a coefficient of kinetic friction of 0.210. Package B has a mass of 11.5kg and a coefficient of kinetic friction of 0.160. How long does it take package A to reach the bottom?

Explanation / Answer

Forces on package A are mag vertically downwards. Friction force Fa acting upwards along the slope. Normal reaction Na from the slope and the force Fab due to block B acting downwards along the slope.

Forces on package B are mbg vertically downwards. Friction force Fb acting upwards along the slope. Normal reaction Nb from the slope and the force Fab due to block A acting upwards along the slope.

Forces mag and mbg can be resolved into directions normal to the slope downwards as mag Cos and mbg Cos and along the slope downwards as mag Cos and mbg Cos respectively.

Therefore, Fa = amagCos and Fb = bmbgCos

Balancing forces in direction along the slope on A, magSin - amagCos + Fab = maaa

Balancing forces in direction along the slope on B, mbgSin - bmbgCos - Fab = mbab

When A and B are in contact, their accelerations aa and ab will be equal.

Hence, adding both the above equations, (ma + mb)gSin - (ama + bmb)gCos = (ma+mb)a

So, a = gSin - (ama + bmb)gCos/(ma+mb)

By putting values, a = 9.81*Sin20 - (0.21*5.5 + 0.16*11.5)*9.81*cos20/(5.5+11.5)

a = 1.73 m/s2

Using s = ut + 1/2*at2 we get

2 = 0 + 1/2*1.73*t2

So, time t = 1.52 s

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