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daser333 [38]
3 years ago
14

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)?
Physics
2 answers:
Ymorist [56]3 years 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]3 years 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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