A block mass 0.5 kg is pushed against a horizontal spring of negligible mass, co

Question

A block mass 0.5 kg is pushed against a horizontal spring of negligible mass, compressing the spring distance (deltaX) the spring constant is 450 N/m. When released the box travels along a frictionless horizontal surface to point B the bottom of a verticle circular track of radius 1meter and continues to move up the track. 1: if the box barely makes it through the top of the loop, what is (deltaX)?
2: what is the speed of the block at the bottom of the loop (v subscript B)? A block mass 0.5 kg is pushed against a horizontal spring of negligible mass, compressing the spring distance (deltaX) the spring constant is 450 N/m. When released the box travels along a frictionless horizontal surface to point B the bottom of a verticle circular track of radius 1meter and continues to move up the track. 1: if the box barely makes it through the top of the loop, what is (deltaX)? 2: what is the speed of the block at the bottom of the loop (v subscript B)?

Explanation / Answer

let VT is speed of the block at the top of the loop not to fall down.

VT^2 = g*R

vT = sqrt(g*R)

now use kinematic equation,

VB^2 - VB^2 = 2*g*H

VB^2 - (g*R) = 2*g*(2*R)

VB^2 = 5*g*R

VB = sqrt(5*g*R)

so, speed of the block at the bottom of the loop,

vB = sqrt(5*g*R)

= sqrt(5*9.8*1)

= 7 m/s <<<<<<<-----------Answer

now use enrgy conservation,

0.5*k*x^2 = 0.5*m*vB^2

x = vB*sqrt(m/k)

= 7*sqrt(0.5/450)

= 0.233 m <<<<<<<-----------Answer

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