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प्रश्न
A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2.
उत्तर
(a) To make the block move up the incline, the applied force should be equal and opposite to the net force acting down the incline.
Applied force = μR + 2g sin 30° (1)
(where μ is the coefficient of static friction)
R = mg cos 30°
Substituting the respective values in Equation (1), we get
`=0.2xx(9.8)sqrt3+2xx9.8xx(1/2)`
3.39 + 9.8 ≈ 13 N
With this minimum force, the body moves up the incline with a constant velocity as the net force on it is zero.
(b) Net force acting down the incline is given by
F = 2g sin 30° − μR
`=2xx9.8xx1/2-3.99`
= 6.41 N
Because F = 6.41 N, the body will move down the incline with acceleration, hence the force required is zero.
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