Question

1. The mass of a cup is 5kg, what is the force acting on the cup g=10m per seconds square

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=40×23

​​=203

​N

While the uertical component of the force acting in upward direction=Fsin30=40×21​=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=203

​N−6N=28.64N

The horizontal acceleration of the block=mFcos30−μN​=528.64​=5.73m/s2

Explanation:

Answer 2

Answer:

Explanation:

Given:  Mass of a cup =5 kg

             Acceleration due to gravity = 10 $m/s^{2}$

             Force = Mass * Acceleration

                         =  5 * 10

                         = 50 $kg m/s^{2}$