Question

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Answer 1

Solution:-When the block tries to move downward,,,the force acting on it,which is stopping it or forcing it to move downward...

1) sliding force i.e mgsin∅(trying to pull it down)

2) spring force i.e kx(acting against the sliding)

3)Normal force which is balanced by mycos∅

These are the total forces which are acting on the block...

Now let us use the The law of conservation of energy to solve this question

since we can assume energy as 0 at any point,,,so,

let us assume potential energy 0 at A

so,the potential energy at B C =mgh=total energy of the block

when,it move downward i e then it's some energy is converted into, kinetic energy ,some used to overcome friction and some energy get stored in spring

so,

the same of this total energy .ust be eqaul to intitial total energy which is mgh

when it moves upto height h

energy stored in spring =1/2 ×k ×x^2=1/2 ×8×0.5=2

used in friction=work done to overcome it=f×s

f=u×N

here N=mgcos37° and u =1/8 (angle CAB=AngleO)

s=0.5

so,

energy used to overcome friction=1/8 ×mg×3/5 ×0.5=1.5mg/40

let the speed final speed be v,,

so kinetic energy=1/2.×mv^2

adding all will be eqaul to mgh

mgh=1/2×mv^2+1.5mg/40 +2

10h=1/2× v^2+1.5/4 +2

also,

h=0.5cos37=0.5×3/5=.3

10×.3=v^2/2 +.375+2

3-2.4=v^2/2

.6×2=v^2

v^2=1.2

v=√1.2

v=.34m/s

hence the velocity of block after sliding upto Distance 0.5 will be 0.34m/s

(hope it helps)