Question

plz answer 3 question

Answer 1

3) you can solve this by many ways but I will used here concept of energy .
here Fext =0
so ,
Change in Kinetic energy = wprdone +change in potential energy

initially when object at top 3m above from ground . { h=√(5²-4²) =3
kinetic energy = 1/2mv² = 1/2m×0 =0
potential energy = -mgh = -mg(3) = -3mg

finally , when body reaches at the bottom
kinetic energy = 1/2mv²
potential energy = -mg×0 = 0

workdone = F.S = mgsin∅ .5
sin∅ = 3/5
so, workdone = 3mg/5

now,

1/2mv² -0 = 3mg/5 + ( 0 -(-3mg))
1/2 v² = 3g/5 + 3g
1/2v² = 18g/5
v² =36×2/5 = 72
v = 6√2 m/sec



2) K.E is directly proportional to square of velocity and mass .
so, mass quadrapted then K.E will be quadrapted .
now,

1/2mv² = 1/2(4m)V²

V² = v²/4

V = v/2

hence , velocity will be half of initial.