Question

A mass m starting from A reaches B of a frictionless track. On reaching A it pushes the track with a force equal to x times its weight, then applicable relation is

• A. h = $\frac { ( x + 5 ) } { 2 } \mathrm { r }$
• B. h =
• C. h = r
• D. h =

Answer 1

Answer: (A)

h = $\frac { ( x + 5 ) } { 2 } \mathrm { r }$

Answer 2

K.E. of block at B = P.E. at A – P.E. at B

i.e., $\frac { 1 } { 2 }$ mv² = mgh – mg²r = mg(h – 2r)

i.e., v² = 2g(h – 2r)………. (i)

Also - mg = x mg

or v² = (x +1)rg ……….. (ii)

Equating (i) and (ii)

2g(h – 2r) = (x +1)gr

or 2gh = (x + 1)gr + 4gr

= (x + 5)gr

or h = $\left( \frac { x + 5 } { 2 } \right) \mathrm { r }$