Download Problem CP2 Chapt 4 - My Solution PDF with thumbnails 2/29/04

Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Chapter4 CP2
Griffith, W. Thomas; The physics of everyday phenomena: a conceptual introduction for
physics;4th Edition ISBN 0-07-250977-5
THE PROBLEM STATEMENT
Ch4 CP2 A rope exerts a constant horizontal force of 250 N to pull a 60-kg crate across the floor. The
velocity of the crate is observed to increase from 1 m/s to 3 m/s in a time of 2 seconds under the influence
of this force and the frictional force exerted by the floor on the crate.
a. What is the acceleration of the crate?
b. What is the total force acting upon the crate?
c. What is the magnitude of the frictional force acting on the crate?
d. What force would have to be applied to the crate by the rope in order for the crate to move with constant
velocity? Explain.
YOU TRY IT HERE FIRST !!!
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Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch4 CP2 A rope exerts a constant horizontal force of 250 N to pull a 60-kg crate across the floor. The
velocity of the crate is observed to increase from 1 m/s to 3 m/s in a time of 2 seconds under the influence
of this force and the frictional force exerted by the floor on the crate.
a. What is the acceleration of the crate?
b. What is the total force acting upon the crate?
c. What is the magnitude of the frictional force acting on the crate?
d. What force would have to be applied to the crate by the rope in order for the crate to move with constant
velocity? Explain.
Definitions, concepts , principles
1.
2.
3.
4.
acceleration a = Δv/Δt
Newton’s 2nd Law of Motion Fnet = ma
Here it is moving all the time, so kinetic friction act all the time.
Friction always acts opposite to the pending or
pending or actual
actual motion, hence opposite to any actively
v velocity (motion)
applied force like the pull on the rope. See the
Fapplied
picture.
m=60 kg
F (friction)
Discussion
The 250 N force is the pull on the rope. It is not the total or net force, Fnet , on the crate.
Friction is the additional force on the block. So, the total or net force, Fnet , on the crate is the pull
on the rope plus th friction force.
m=60 kg
F (friction)
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a
Frope
Physics 110 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Basic Solution (Minimum Expected from the student)
Ch4 CP2 A rope exerts a constant horizontal force of 250 N to pull a 60-kg crate across the floor. The
velocity of the crate is observed to increase from 1 m/s to 3 m/s in a time of 2 seconds under the influence
of this force and the frictional force exerted by the floor on the crate.
a. What is the acceleration of the crate?
b. What is the total force acting upon the crate?
c. What is the magnitude of the frictional force acting on the crate?
d. What force would have to be applied to the crate by the rope in order for the crate to move with constant
velocity? Explain.
a. acceleration a = Δv/Δt
to quote the information directly given “The velocity of the crate is observed to increase from
1 m/s to 3 m/s in a time of 2 seconds, “
So,
acceleration a = Δv/Δt = (v2 - v1)/ Δt = (3m/s - 1 m/s)/2s = (2m/s)/2s = 1 m/s2
b. Newton’s 2nd Law of Motion Fnet = ma.
So, Fnet = ma = 60 kg * 1 m/s2 = 60 (kg*m/s2) = 60 N .
c. The total force or net force Fnet on the crate is 250 N pull on the rope plus the friction Ffriction
Since friction always acts opposite to the pending or actual motion, hence opposite to any
actively applied force like the pull on the rope. Then
Fnet = 250N pull on the rope - Ffriction .
So,
Ffriction = 250 N pull on the rope - Fnet = 250 N - 60 N = 190 N.
The math
Fnet = 250N pull on the rope - Ffriction
-Fnet
-Fnet
to get rid of Fnet on left side of equation
___________________________________ subtract Fnet from both sides.
0 = 250N pull on the rope - Ffriction - Fnet Gives the result.
Ffriction
to get rid of - Ffriction of both sides of the
Ffriction
__________________________________
equation, add Ffriction from both sides.
Ffriction = 250 N pull on the rope - Fnet
Gives the result.
d. To keep the crate moving at a constant velocity Newton’s 2nd Law of Motion Fnet = ma,
requires that the acceleration a = 0. So, Fnet = ma = 0.
From the picture,
a=0
F
Frope = Ffriction = 190 N
m=60 kg
rope
F (friction)
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