A 1.1 kg block slides on a horizontal frictionless surface in a circular path at the end of a 0.50 m long string. (a) Calculate the block’s speed if the tension in the string in 86 N. (b) By what percent does the tension change if the block speed decreases by 10 percent?

Explanation

(a) Force diagram for the block is shown in fig. The upward normal force balances the block’s weight. The tension force of the string on the block provides the centripetal force that keeps the block moving in a circle. Newton’s second law for forces along the radial direction is `sum` F
(in radial direction ) = T = `(mv^(2))/(r ) `.
or v = `sqrt((Tr)/(m)) = sqrt(((80N)(0.50m))/(1.2 kg))` = 5.0 m/s
(b) A 10 percent reduction in the speed results in a speed v’ = 5.4 m/s. The new tension is
T' = `(mv^(2))/(r ) = ((1.2kg)(5.4m//s)^(2))/(0.50m)` = 70N
thus, `(T')/(T) = (70N)/(86N) = 0.81`
percentage reduction in the tension is about 19%. The same result is obtained using a proportionality method.
`(T')/(T) = ((mv^('2)//r))/((mv^(2)//r)) = ((v')/(v))^(2) = ((0.90)/(v))^(2) = 0.81`