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 hori

Question

3 years ago

10

zontal. The magnitude of the friction force acting on the block is:

1 answer:

Kamila [148] · Answer 1 · 2022-01-26T04:39:20+03:00

The force of friction is 25 N

Explanation:

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

$\sum F_x = ma_x$

where

$\sum F_x$ is the net force in the horizontal direction

m is the mass of the block

$a_x$ is the horizontal acceleration

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

$a=0 \rightarrow \sum F_x = 0$ (1)

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

$\sum F_x = Fcos \theta -F_f$ (2)

where

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

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

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

Learn more about friction:

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