Question

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)?

Answer 1

The magnitude of this horizontal force can be calculated as :

F=mg
2x9.8=19.6N
19.6cos 30= 17 Newtons

Answer 2

Horizontal force is 16.97 N.

Step-by-step 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* 9.8\ m/s^2\ sin(60)`

`F_x=16.97\ N`

So, the horizontal component of force is 16.97 N.