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kirill115 [55]
3 years ago
5

Only two forces act on an object (mass = 3.80 kg), as in the drawing. (F = 51.0 N.) Find the magnitude and direction (relative t

o the x axis) of the acceleration of the object.

Physics
1 answer:
natita [175]3 years ago
4 0
F x = F · cos 45° = 51 N · 0.7071 = 36.1 N
F y = F · sin 45° = 51 N  · 0.7071 = 36.1 N
Q = m g = 3.8 kg · 9.81 m/s² = 29.43 N
R y = 36.1 N - 29.43 N = 6.67 N
R x = 40 N + 36.1 N = 76.1 N
R = √ (R x² + R y²) = √ ( 44.4889 + 5791.21 ) = 76.39 N
F = m a
a = F/ m
a = 76.39 N: 3.8 kg = 20.1 m/s²
The magnitude of acceleration is 20.1 m/s².
α = tan ^(-1) ( R y / R x ) = 6.67 / 7.61 = 0.0876478
α = 5°
Direction is α = 5° counterclockwise from the + x-axis.
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in Above equation in x and t. Separating the variables and integrating,

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.

.

.

.

.

.

.

.

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Here

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      = −

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e

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l

/

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