A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a
t an angle of 23o from the horizontal. The floor is rough and has a coefficient of kinetic equal to 0
a) (2 pts.) Model: How will you model this problem?
b) (5 points) Visualize: Draw a free body diagram showing all forces acting on the crate. Be sure to label the direction of your positive unit vectors c) (5 points)
c) Visualize: Draw a pictorial representation of the crate and label all known quantities
d) (5 pts.) Solve: Calculate the work done by the gravitational force e) (15 pts.)
e) Solve: Calculate the normal force f (5 pts.)
f) Solve: Calculate the work done by the normal force g (10 pts.)
g) Solve: Calculate the magnitude of the frictional force h) (15 pts.)
h) Solve: Calculate the work done by the frictional force. i) (15 pts.)
i) Solve: Calculate the work done by the force exerted by the j) (3 pts.)
j) Solve: Calculate the net work done on the crate. _ person
a) (2 pts.) Model: How will you model this problem?
b) (5 points) Visualize: Draw a free body diagram showing all forces acting on the crate. Be sure to label the direction of your positive unit vectors c) (5 points)
c) Visualize: Draw a pictorial representation of the crate and label all known quantities
d) (5 pts.) Solve: Calculate the work done by the gravitational force e) (15 pts.)
e) Solve: Calculate the normal force f (5 pts.)
f) Solve: Calculate the work done by the normal force g (10 pts.)
g) Solve: Calculate the magnitude of the frictional force h) (15 pts.)
h) Solve: Calculate the work done by the frictional force. i) (15 pts.)
i) Solve: Calculate the work done by the force exerted by the j) (3 pts.)
j) Solve: Calculate the net work done on the crate. _ person
1 answer:
Answer:
Check attachment for solution and diagrams
Explanation:
Given that,
Mass of crate m=63kg
Distance travelled d=40m
Horizontal force Fx=130N
Angle the force applied on cord makes with horizontal is θ=23°.
The weight of the crate is given by
W=mg
W=63×9.81
W=618.03N
Horizontal force Fx=130N
Resolving the applied Force F to the horizontal will give
Fx=FCos θ
F=Fx/Cos θ
F=130/Cos23
F=141.2N
a. Check attachment for model diagram
b. Check attachment for free body diagram
c. Check attachment for pictorial representation
d. Work done by gravitational force.
We, know that the body did not move upward, then the distance d=0
Work done is given as
W=F×d
So, d=0
W=F×0
W=0J
So, no work is done by gravity
e. Normal force?
Using newton law of motion
ΣFy = may
Since the body is not moving upward, then ay=0m/s²
N+141.2Sin23-618.03=0
N=618.03-141.2Sin23
N=562.86N
f. Work done by normal force.
The body is not moving upward, then the distance is zero
d=0
Work done by normal=normal force × distance
Wn=562.86×0
Wn=0J
No work is done by the normal force
g. Frictional force?
Since the coefficient of kinetic friction is zero, then the surface is frictionless
So, no frictional force is acting on the body
Fictional force is given as
Fr=μk•N
Given that, μk=0
Fr=0×562.86
Fr=0N
d. Work done by frictional force?
Since the frictional force Is zero, then, no work is done by friction
W(friction ) = frictional force × d
Here, the body moved a distance of 40m
W(fr)=0×40
W(fr)=0J
No work is done by friction
I. Work done by exerted force
The horizontal component of the exerted force is 130N and the body traveled a distance of 40m
Then, work done is given as
Workdone=force ×distance
Work done=130×40
W=5200J
W=5.2KJ
h. Net workdone?
Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.
Went, = work done by force exerted - work done by friction
Wnet=5200-0
Wnet, =5200
Wnet=5.2KJ
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The correct answer is:
Work is negative, the environment did work on the object, and the energy of the system decreases.
In fact, the work-energy theorem states that the work done by the system is equal to its variation of kinetic energy:
In this problem, the variation of kinetic energy is negative (because the final velocity is less than the initial velocity), so the work is negative, and this means that the environment did work on the object, and its energy decreased.