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6. For each situation below, a person pushes a 20 kg crate along a horizontal surface with friction. Assume g = 10 m/s² and complete the following for each case: a) Use the scale provided and determine the magnitude of the normal force on the crate. b) Determine the magnitude of the kinetic friction force assuming the coefficient of kinetic friction is 0.2. c) Use the scale provided and determine the magnitude of the net force in the x direction. d) Determine the acceleration in the x direction. e) Apply Newton's laws in component form using symbols and numbers shown in the diagrams, (I) System y N (?) F_k (?) P (224 N) x F_G (200 N) N = F_k = ?F_x = a_x = (II) y N (?) x F_k (?) P (224 N) F_G (200 N) N = F_k = ?F_x = a_x = (III) y N (?) x F_k (?) F_G (200 N) P (224 N) N = F_k = ?F_x = a_x =

          6. For each situation below, a person pushes a 20 kg crate along a horizontal surface with friction. Assume g = 10 m/s² and complete the following for each case:
a) Use the scale provided and determine the magnitude of the normal force on the crate.
b) Determine the magnitude of the kinetic friction force assuming the coefficient of kinetic friction is 0.2.
c) Use the scale provided and determine the magnitude of the net force in the x direction.
d) Determine the acceleration in the x direction.
e) Apply Newton's laws in component form using symbols and numbers shown in the diagrams,
(I)
System
y
N (?)
F_k (?)
P (224 N)
x
F_G (200 N)
N =
F_k =
?F_x =
a_x =
(II)
y
N (?)
x
F_k (?)
P (224 N)
F_G (200 N)
N =
F_k =
?F_x =
a_x =
(III)
y
N (?)
x
F_k (?)
F_G (200 N)
P (224 N)
N =
F_k =
?F_x =
a_x =

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6. For each situation below, a person pushes a 20 kg crate along a horizontal surface with friction. Assume g = 10 m/s² and complete the following for each case:
a) Use the scale provided and determine the magnitude of the normal force on the crate.
b) Determine the magnitude of the kinetic friction force assuming the coefficient of kinetic friction is 0.2.
c) Use the scale provided and determine the magnitude of the net force in the x direction.
d) Determine the acceleration in the x direction.
e) Apply Newton's laws in component form using symbols and numbers shown in the diagrams,
(I)
System
y
N (?)
Fk (?)
P (224 N)
x
FG (200 N)
N =
Fk =
?Fx =
ax =
(II)
y
N (?)
x
Fk (?)
P (224 N)
FG (200 N)
N =
Fk =
?Fx =
ax =
(III)
y
N (?)
x
Fk (?)
FG (200 N)
P (224 N)
N =
Fk =
?Fx =
ax =

Added by Jesse H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Stephen Zaffke

07/15/2022

Step 1

2 * 200 N = 40 N c) Net force in the x direction = Pushing force - Friction force = 224 N - 40 N = 184 N d) Acceleration in the x direction = Net force / Mass = 184 N / 20 kg = 9.2 m/s^2 ** Show more…

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For each situation below, a person pushes a 20 kg crate along a horizontal surface with friction. Assume g = 10 m/s² and complete the following for each case: a) Use the scale provided and determine the magnitude of the normal force on the crate. b) Determine the magnitude of the kinetic friction force assuming the coefficient of kinetic friction is 0.2. c) Use the scale provided and determine the magnitude of the net force in the x direction. d) Determine the acceleration in the x direction. e) Apply Newton's laws in component form using symbols and numbers shown in the diagrams,

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Transcript

00:01 Okay, so the first case is the simplest case.

00:03 We have a pushing force that is directly horizontal, and we have a normal force, a friction force, and a weight force.

00:11 We know the mass and the coefficient of friction, and we know the pushing force is 224 newtons.

00:17 So in that first case with the horizontal pushing force, the normal force that it will feel will directly balance the weight force.

00:30 So the weight force being 200 newtons means that the normal force will also be 200 newtons.

00:37 And the friction force is going to be mu times that normal force.

00:42 So if the normal force is 200, we multiply that by 0 .2, and we get a friction force of 40 newtons.

00:51 That should be newtons there.

00:56 Let's make that clear clearly an n.

00:59 There we go.

01:01 And then for the net force, we have a pushing force to the right and a friction force to the left.

01:11 So the difference between those is 184 newtons.

01:14 And the acceleration that it will feel will be whatever that net force is divided by the mass.

01:20 So 1804 divided by 20, we get 9 .2 meters per second square.

01:25 So that's case one.

01:27 Case two, we are now pushing at an angle.

01:30 So our angle is going to look like this.

01:33 So we're going to have an x component and a y component.

01:37 And this angle, we can tell because the vector goes two blocks to the right and one block down.

01:45 So a tangent of that angle is 1 over 2...

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