Varying Force on Sliding Crate

Question:

A force acts horizontally on a 50 pound crate sitting at rest on the floor. The magnitude of the force is 2.4*t² pounds where t is time in seconds. The coefficient of static friction between the floor and the crate is 0.3. The coefficient of kinetic friction between the floor and the crate is 0.2. Determine the speed of the crate at time t=5 seconds.

Answer:

The force required to get the crate moving is the normal force holding the crate to the floor times the coefficient of static friction, 50*0.3 pounds = 15 pounds. Until the applied force has reached that value, the crate does not move. When 2.4*t²=15, t=2.5 seconds. At that time the crate starts to move. At any instant after that its acceleration is the difference between the applied force and the force of kinetic friction, divided by the mass of the crate. The frictional force is the normal force times 0.2 or 10 pounds. The initial acceleration then is (15-10)/1.553=3.22f/s².

The acceleration at any time after 2.5 seconds then is force at that time minus 10 pounds for friction divided by 1.553 slugs plus the intial acceleration. In equation form acceleration
a=(2.4*t²-10)/1.553 + 3.22= 1.5454*t²-3.22.
The velocity at time t=5 seconds is the integral of acceleration over time,
1.5454/3*t^3-3.22*t,
evaluated from t=2.5 to t=5.
Remembering that up to t=2.5 there is no acceleration the integral evaluates to:
1.5454/3*5^3-3.22*5 - 1.5454/3*2.5^3 =
64.392 - 16.1 - 8.05 = 40.242 f/s.

This information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know.
JDJ