Answer
(a) T = 0.074 N
(b) T = 4.7 N
(c) The person did 0.22 J of work.
Work Step by Step
In this situation, the tension provides the centripetal force to keep the object moving in a circle.
(a) $T = \frac{mv^2}{r}$
$T = \frac{(0.0600~kg)(0.70~m/s)^2}{0.40~m}$
$T = 0.074~N$
(b) $T = \frac{mv^2}{r}$
$T = \frac{(0.0600~kg)(2.80~m/s)^2}{0.10~m}$
$T = 4.7~N$
(c) $W = K_2-K_1$
$W = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$
$W = \frac{1}{2}(0.0600~kg)(2.80~m/s)^2-\frac{1}{2}(0.0600~kg)(0.70~m/s)^2$
$W = 0.22~J$
The person did 0.22 J of work.
