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When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, the magnitude of the normal force of the floor on the crate isA. 25N
B. 68N
C. 180N
D. 250N
E. 310N

Respuesta :

Kazeemsodikisola

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

Ver imagen Kazeemsodikisola
onyebuchinnaji

The magnitude of the normal force of the floor on the crate is 180 N.

The given parameters;

  • mass of the crate, m= 25 kg
  • frictional force, F = 200 N
  • angle of inclination, = 20⁰

The magnitude of the normal force of the floor on the crate is calculated as follows;

[tex]F_n = W - F_f\\\\F_n = (mg) - (F sin \ \theta )\\\\F_n = (25 \times 9.8) - (200 \times sin20)\\\\F_n = 176.6 \ N \\\\F_n \approx 180 \ N[/tex]

Thus, the magnitude of the normal force of the floor on the crate is 180 N.

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