The curved portion is smooth and the horizontal portion are rough.The block is released from P.At what distance from A the block will stop. ( μ=0.2) ?

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2 Answers
Feb 25, 2018

#1 m#

Explanation:

During the fall from #1 m# height,the block will gain kinetic energy and the total energy will be purely kinetic on coming to point #A#,which was purely potential energy while sitting on top of the curve.

Applying law of conservation of energy,

we can say, #mgh = 1/2mv^2# (where, #v# is the velocity at #A#)

So, #v^2=19.62#

Now,the frictional force that will be acting on the block is #mu N=mu mg# (#mu# = coefficient of friction=#0.2#)

So,deceleration while going along the horizontal part will be #(mu mg)/m = mug=1.962 ms^-2#

So,if it stops by going distance #x# then applying, #v^2=u^2 - 2ax#

we get, #0^2 = 19.62 - 2*1.962 x#

So, #x=5 m#

But,givn length of the horizontal part is #2 m#, so let us see what will be the velocity,after travelling through #2m#,

So, #v'^2 = 19.62 - 2*1.962 *2=11.76#

So,at point #B# total energy = kinetic energy = #1/2m v'^2#

Now,if it goes up by height #s# then, again applying law of conservation of energy,

# 1/2 mv'^2 = mgs#

So, #s=0.6 m#

Now,after reaching that point it will again come down and gain the same velocity at point #B# while it was going up,so now if it moves distance #r# towards #A#, we can check like previously,

#0^2 = 11.76 - 2*1.962 *r#

So, #r=3#

So,it will still not come to rest still now, so lets find velocity after moving #2m# to reach at #A#

so, #u^2 = 11.76 - 2*1.962*2=4 #

So,after rising up when coming down,it will have same velocity at #A#,

So,this time if it stops after going #ym#,we can say,

#0^2 =4 - 2*1.962*y#

or, #y=1.01#

Almost about #1 m# from #A# after these above set of rides the block will stop.

Feb 26, 2018

1m from A

Explanation:

The initial total energy is just the potential energy : #mgh# - where #h# is given as 1 m. For the body to come to a complete halt, this energy has to be dissipated in the form of work done by friction against the motion of the block. Since the force of friction is given by #mu mg# in magnitude, the distance that the block has to slide over the rough ground is #{mgh}/{mu mg} = h/mu = 5# m

Thus the block will cover the 2m stretch of rough ground twice and then travel an extra distance of 1 m before coming to a halt (in between it will climb up and down the two curved surfaces, but since these are smooth, this will not dissipate any energy ).