text: does cathy's cake slide forward? a taxi suddenly cuts in front of the bus, causing the bus drivere to slam on the breaks. the bus driver reduces speed from 60.0km/h to 20.0km/h in 1.5 seconds. the static friction is 1.0, the mass of the cake is 2.0kg. does cathy's cake slide forward? 60km/hr = 16.666m/s 20km/hr = 5.555m/sc  abus = (5.555m/s-16.666m/s)/1.5s = -7.4073m/s^2   fnet = fbus - ffriction = fcake fbus = m*abus fcake = m*acake m*abus - ffriction = fcake m*abus - u*fn = fcake m*abus - us*m*g = fcake (2.0kg)(-7.4073m/s^2) - (1.0)(2.0kg)(9.8m/s^2)=fcake fcake= -34.42n |fcake| = 34.42n  i took the absolute value because we'll be dealing with magnitudes.  in order for motion to occur, the force of the cake must exceed the maximum force of static friction.  ffmax = us*fn = us*m*g ffmax = (1.0)(2.0kg)(9.8m/s^2) = 19.6n and  fcake = 34.42n   ffmax < fcake  because the cake's force is greater than that of the maximum static force, the cake will move.  **note: i'm on very little sleep so hopefully this is correct and makes sense**