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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Answer:
His launching angle was 14.72°
Explanation:
Please, see the figure for a graphic representation of the problem.
In a parabolic movement, the velocity and displacement vectors are two-component vectors because the object moves along the horizontal and vertical axis.
The horizontal component of the velocity is constant, while the vertical component has a negative acceleration due to gravity. Then, the velocity can be written as follows:
v = (vx, vy)
where vx is the component of v in the horizontal and vy is the component of v in the vertical.
In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):
sin angle = opposite / hypotenuse
cos angle = adjacent / hypotenuse
In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:
sin angle = v0y / v0 then: v0y = v0 * sin angle
In the same way for vx:
vx = v0 * cos angle
Using the equation for velocity in the x-axis we can find the equation for the horizontal position:
dx / dt = v0 * cos angle
dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)
x - x0 = v0 t cos angle
x = x0 + v0 t cos angle
For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:
dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)
vy -v0y = g t
vy = v0y + g t
vy = v0 * sin angle + g t
The position will be:
dy/dt = v0 * sin angle + g t
dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)
y = y0 + v0 t sin angle + 1/2 g t²
The displacement vector at a time "t" will be:
r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)
If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)
r = (x0 + v0 t cos angle, 0)
The module of this vector will be the the total displacement (65 m)
module of r =
65 m = x0 + v0 t cos angle ( x0 = 0)
65 m / v0 cos angle = t
Then, using the equation for the position in the y-axis:
y = y0 + v0 t sin angle + 1/2 g t²
0 = y0 + v0 t sin angle + 1/2 g t²
replacing t = 65 m / v0 cos angle and y0 = 0
0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²
cancelating v0:
0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)
-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)
using g = -9.8 m/s²
-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²
sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²
(using trigonometric identity: sin x cos x = sin (2x) / 2
sin (2* angle) /2 = 0.25
sin (2* angle) = 0.49
2 * angle = 29.44
<u>angle = 14.72°</u>
