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A 925 N crate is being pulled across a level floor
by a force F of 325 N at an angle of 25 above the
horizontal. The coefficient of kinetic friction is
0.25. Find the magnitude of the acceleration of
the crate.

Respuesta :

pfellas
The normal contact force= 925 N
The friction = 0.25 x 925 = 231.25 N
The horizontal component of the force = 925 x Cos 25 = 838.3 N
The net force = 838.3 - 231.25 = 607 N

acceleration = 607 / (925/9.8) = 6.4 m/s^2
onyebuchinnaji

The magnitude of the acceleration of  the crate is 1.033 m/s².

The given parameters;

  • weight of the crate, W = 925 N
  • horizontal force applied to the carte, F = 325 N
  • angle of inclination of the force, θ = 25⁰
  • the coefficient of friction, μ = 0.25

The mass of the crate is calculated as;

W = mg

[tex]m = \frac{W}{g} \\\\m = \frac{925}{9.8} \\\\m = 94.388 \ kg[/tex]

The normal force on the crate is calculated as;

Fₙ = 925 - 325 x sin(25)

Fₙ = 925 - 137.35

Fₙ = 787.65 N

The frictional force on the object is calculated as follows;

[tex]F_k = \mu F_n\\\\F_k = 0.25 \times 787.65\\\\F_k = 196.91 \ N[/tex]

The magnitude of the crates acceleration is calculated from the net horizontal force on the crate;

[tex]\Sigma F_x = 0\\\\Fcos (\theta) - F_k = ma\\\\325\times cos(25) \ - \ 196.91 = ma\\\\97.54 = ma[/tex]

[tex]a = \frac{97.54}{94.388} \\\\a = 1.033 \ m/s^2[/tex]

Thus, the magnitude of the acceleration of  the crate is 1.033 m/s².

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