Question

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:

Answer 1

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

• The horizontal component of the pulling force, $F cos \theta$, where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the  forward direction
• The force of friction, $F_f$, acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$

where

m is the mass of the block

$a_x$ is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

$a_x = 0$

So the equation becomes

$\sum F_x = 0$ (1)

The net force here is given by

$\sum F_x = F cos \theta - F_f$ (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

$F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N$

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