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A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the horizontal. The magnitude of the friction force acting on the block is:

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skyluke89

The force of friction is 25 N

Explanation:

For this problem, we can apply Newton's second law of motion along the horizontal direction:

[tex]\sum F_x = ma_x[/tex]

where

[tex]\sum F_x[/tex] is the net force in the horizontal direction

m is the mass of the block

[tex]a_x[/tex] is the horizontal acceleration

Here  the block is moving at constant speed, so its acceleration is zero, therefore:

[tex]a=0 \rightarrow \sum F_x = 0[/tex] (1)

The net force in the horizontal direction can be written as:

[tex]\sum F_x = Fcos \theta -F_f[/tex] (2)

where

  • [tex]Fcos\theta[/tex] is the horizontal component of the pulling force, with F = 50 N being the magnitude and [tex]\theta=60^{\circ}[/tex] being the direction, acting forward
  • [tex]F_f[/tex] is the force of friction, acting backward

Combining (1) and (2), we find the magnitude of the force of friction:

[tex]Fcos \theta -F_f = 0\\F_f = F cos \theta =(50)(cos 60^{\circ})=25 N[/tex]

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