Question

While slipping on rough spherical surface of radius 'R', block A of mass 'm' comes with velocity $\sqrt { 1.4 \mathrm { gR } }$ at bottom B. Work done in slipping the block from 'B' to 'C' is –

• A. $\frac { \mathrm { mgR } } { 4 }$
• B. mgR
• C. 1.3 mgR
• D. $\frac { 5 } { 4 }$ mgR

Answer 1

Answer: (C)

1.3 mgR

Answer 2

Loss in P.E. = gain in K.E. + work done against friction mg R = $\frac { 1 } { 2 }$ m (1.4 gR) + W_f

W_f = 0.3 mgR

Now

W _{B ࢮ C} = mg R + 0.3 mgR

= 1.3 mgR