Question

A block of 5kg is placed on a smooth horizontal surface. A variable Force F=2t Where t is time in second is acting on the block at an angle 60 degree with vertical .The time when the block will leave the contact from the surface is (g=10m/s2)
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Answer 1

Answer:

Mass of the block=5kg

Coeffecient of friction=0.2

external applied force, F=40N

The angle at which the force is applied=30degree

So the horizontal component of force=Fcos30=40×

2

3

=20

3

N

While the uertical component of the force acting in upward direction=Fsin30=40×

2

1

=20N

The normal reaction from the surface (N)=mg−Fsin30=50−20=30N

So the ualue of limiting friction=μN=0.2×30=6N

Hence the net horizontal force on the block=Fcos30=μN=20

3

N−6N=28.64N

The horizontal acceleration of the block=

m

Fcos30−μN

=

5

28.64

=5.73m/s

2