Q. A block of mass 0.5 kg rests ona horizontal surface when a horizontal of force 2.0 newton on it it acquires an acceleration of 3m/sec square the force of friction between the block and the horizontal surface is ?
PLZZ HELP ME......................!!
Answers
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=203N
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=203N−6N=28.64N
The horizontal acceleration of the block=mFcos30−μN=528.64=5.73m/s2
Answer:
hi buddy happy lohri
Step-by-step explanation:
ans Maine apko 2nd project ka bataya tha na
Pahala wala project is completed and 2nd wala
AR house built program h
Uske liye apka full address Chahiye padegaa ( if you don't mind , worry I never mis use this ok) with area Pincode and ho sake to any landmark too
Apke house ki image BHI Chahiye padegi
aap apna email bata dena toh complete hone pr apko pic send kr dunga