Question

: A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks

Answer 1

Answer:

r = 0.0173 m = 1.73 cm

Explanation:

Here, the centripetal force of the block will be providing the required breaking tension in the string:

$Tension = Centripetal Force\\T = F_c\\\\T = \frac{mv^2}{r} \\\\r = \frac{mv^2}{T}$

where,

r = radius = ?

m = mass of block = 0.13 kg

v = tangential spee of block = 4 m/s

T = Breaking Strength = 30 N

Therefore,

$r = \frac{(0.13\ kg)(4\ m/s)^2}{30\ N}$

r = 0.0173 m = 1.73 cm