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*t^{2} 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*t^{2}=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/s^{2}.
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*t^{2}-10)/1.553 + 3.22=
1.5454*t^{2}-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.

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