Question

A 10 N block is released from rest at A and slides down along the smooth cylindrical surface. The attached spring has a stiffness,
k = 30 Nm^–1. If it does not allow the block to leave the surface until q = 60º, its unstretched length is –

• A. 0.166 m
• B. 0.5 m
• C. 0.333 m
• D. 0.666 m

Answer 1

Answer: (C)

0.333 m

Answer 2

N₁ = mg + kx₁ …(1)

mg cos 60⁰ – N₂ + kx₂ = $\frac { \mathrm { mv } ^ { 2 } } { \mathrm { R } }$

Block will leave surface if N₂ = 0

$\frac { \mathrm { mg } } { 2 }$ + kx₂ = mg $\frac { \mathrm { v } ^ { 2 } } { \mathrm { R } }$ …(2)

Conserving energy at A & B

$\frac { 1 } { 2 }$ mv² = mgR(1 – cos 60⁰) = $\frac { \mathrm { mgR } } { 2 }$

v = $\sqrt { \mathrm { gR } }$

from (2) x₂ = $\frac { \mathrm { mg } } { 2 \mathrm { k } }$ = 0.166 m

Unstretched length x₀ = R – x₂

= 0.5 – 0.166 = 0.333 m