Answer
The radial acceleration component is $2.45~m/s^2$
The tangential acceleration component is $2.54~m/s^2$
The tension in the string is $11.9~N$
Work Step by Step
We can find the radial acceleration component:
$a_c = \frac{v^2}{r}$
$a_c = \frac{(1.40~m/s)^2}{0.800~m}$
$a_c = 2.45~m/s^2$
The radial acceleration component is $2.45~m/s^2$
We can find the tangential acceleration component:
$mg~sin~\theta = ma_t$
$a_t = g~sin~\theta$
$a_t = (9.80~m/s^2)~sin~15.0^{\circ}$
$a_t = 2.54~m/s^2$
The tangential acceleration component is $2.54~m/s^2$
We can consider the net force on the bob directed along the line toward the center of the circle. We can find the tension in the string:
$\sum~F = \frac{mv^2}{r}$
$T - mg~cos~\theta = \frac{mv^2}{r}$
$T = mg~cos~\theta + \frac{mv^2}{r}$
$T = (1.00~kg)(9.80~m/s^2)~cos~15.0^{\circ} + \frac{(1.00~kg)(1.40~m/s)^2}{0.800~m}$
$T = 11.9~N$
The tension in the string is $11.9~N$
