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1 year ago

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a 2.0 kg block on an incline at a 60.0 degree angle is held in equilibrium by a horizontal force, what is the magnitude of this

horizontal force (disregard friction)?

2 answers:

Ymorist [56]1 year ago

8 0

Answer:

Horizontal force is 16.97 N.

Explanation:

It is given that,

Mass of the block, m = 2 kg

It is on an incline at a 60 degree angle is held in equilibrium by a horizontal force. We need to find the magnitude of horizontal force. The force acting on the block is its weight mg.

The horizontal component of force is, $F_x=mg\ sin\theta$

The vertical component of force is, $F_y=mg\ cos\theta$

So, the horizontal component of force is, $F_x=2\ kg\times 9.8\ m/s^2\ sin(60)$

$F_x=16.97\ N$

So, the horizontal component of force is 16.97 N.

fiasKO [112]1 year ago

3 0

<span>The magnitude of this horizontal force</span> can be calculated as :

F=mg
2x9.8=19.6N
<span>19.6cos 30= 17 Newtons
</span>

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<em>Comment on the units</em>

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