3. A crate is pushed in the positive direction along a rough, horizontal surface with a force Fthat makes an angle e 45.00 below the horizontal. The magnitude of the force applied isF-200. N. The mass of the crate is m 30.0 kg. The coefficient of kinetic friction between thecrate and the surface is 0.300.a) Draw a clearly labeled free-body diagram for the crate. Carefully identifying the origin ofeach force. (example: Fe (Earth pulling block))b) Find the acceleration of the crate

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Asked Oct 29, 2019
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3. A crate is pushed in the positive direction along a rough, horizontal surface with a force F
that makes an angle e 45.00 below the horizontal. The magnitude of the force applied is
F-200. N. The mass of the crate is m 30.0 kg. The coefficient of kinetic friction between the
crate and the surface is 0.300.
a) Draw a clearly labeled free-body diagram for the crate. Carefully identifying the origin of
each force. (example: Fe (Earth pulling block))
b) Find the acceleration of the crate
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3. A crate is pushed in the positive direction along a rough, horizontal surface with a force F that makes an angle e 45.00 below the horizontal. The magnitude of the force applied is F-200. N. The mass of the crate is m 30.0 kg. The coefficient of kinetic friction between the crate and the surface is 0.300. a) Draw a clearly labeled free-body diagram for the crate. Carefully identifying the origin of each force. (example: Fe (Earth pulling block)) b) Find the acceleration of the crate

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Expert Answer

Step 1

Given values:

F = 200 N

m = 30.0 kg

the coefficient of kinetic friction between the crate and the surface = 0.3

Step 2

(a)

The free diagram for the crate:

F sin450
R
a
450
F cos450
m
V
mg
ed
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F sin450 R a 450 F cos450 m V mg ed

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Step 3

(b)

Applying the Newton’s second law Along vertical component:

There is ...

R-F sin 45 - mg = 0
_
R mgFsin 45°
Frictional force:
mgFsin 45°)
(i)
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R-F sin 45 - mg = 0 _ R mgFsin 45° Frictional force: mgFsin 45°) (i)

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