Question

4.
A block of mass M = 2 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When
a force F = 20 N is applied, the acceleration of the block will be (g = 10 m/s2)
(1) (1+5√3)m/s2
(2) 53 m/s2
(3) 5 m/s2
(4) (5√3 – 1) m/s2
ans is option 4..plz see the image for reference. thanks​

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

Explanation:

I hope it will help you

Answer 2

Answer:

actually the snswer is 5root3-2 not 5root3-1

Explanation:

please convey if correct mark as brainloest if

not give 1 star and report if the answer

is 4 th option

then mu should be 0.1