General Physics with Calc 1
8. A package of dishes (mass 50.0 kg) sits on the flatbed of a
pickup truck with an open tailgate. The coefficient of static
friction between the package and the truck's flatbed is 0.310, and
the coefficient of kinetic friction is 0.240. (a) The truck
accelerates forward on level ground. What is the maximum
acceleration the truck can have so that the package does not slide
relative to the truck bed?
(b) The truck barely exceeds this acceleration and then moves with
constant acceleration, with the package sliding along its bed. What
is the acceleration of the package relative to the ground?
(c) The driver cleans up the fragments of dishes and starts over
again with an identical package at rest in the truck. The truck
accelerates up a hill inclined at 9.0° with the horizontal. Now
what is the maximum acceleration the truck can have such that the
package does not slide relative to the flatbed?
(d) When the truck exceeds this acceleration, what is the
acceleration of the package relative to the ground?
(e) For the truck parked at rest on a hill, what is the maximum
slope the hill can have such that the package does not slide?
_____°
(f) Is any piece of data unnecessary for the solution in all the
parts of this problem? Pick one below
The coefficient of kinetic friction
The mass of the package
The coefficient of static friction
8. a)
N - mg = 0
N = mg
maximum static friction =us N = 0.310mg
this is the maximum force on package.
and Fnet = ma
0.310mg = ma
a =0.310 x 9.8 = 3.04 m/s^2
b) when package is sliding then kinetic friction will work.
kinetic friction= uk N = 0.240mg
a = F/m = 0.240g = 0.240 x9.8 = 2.35 m/s^2
c) now N - mgcos9 = 0
N = mgcos9
maximum static friction =us N = 0.310mg cos9
this is the maximum force on package.
and Fnet = ma
0.310mgcos9 - mgsin9 = ma
a =9.8(0.310cos9 - sin9) = 1.47 m/s^2
d) when package is sliding then kinetic friction will work.
kinetic friction= uk N = 0.240mg cos9
0.240mgcos9 - mgsin9 = ma
a =9.8(0.240cos9 - sin9) = 0.79 m/s^2
e) at slope @,
N = mgcos@
f = 0.31mgcos@
then along the incline,
mgsin@ - 0.31mgcos@ = 0
tan@ = 0.31
@ = 17.22 deg
f) mass is not neessary.
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