Question

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/s)
(Force is acting at an angle of 30° from horizontal)

(1) (1+5./3mis?
(2) 53 m/s2
(3) 5 m/s2
(4) (513 - 1) m/s?​

Answer 1

Answer:

Explanation::

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