Answer:
0.8582 m/s^2
Explanation:
mass of the bob m = 80 kg
length of the string L= 10.0 m
angle made with the vertical θ= 5.0
let the force exerted by the string = T
from the FBD
T cosθ= mg
T cos 6°= 80×9.81
⇒T= [tex]\frac{70\times9.81}{cos5}[/tex]
= 787.79 N
how horizontal component of tension [tex]T_h[/tex]
[tex]T_h= Tsin\theta[/tex]
=787.79 sin 5°= 68.66 N
Now, radial acceleration, [tex]a_c= \frac{T_h}{m}[/tex]
= 68.66/80= 0.8582 m/s^2