Question

A pendulum consists of a 2.2 kg bob attached to a 2.8 meter long string. The bob has a velocity
of 4.7 m/s as it passes the bottom. Considering the point where the string makes a 29° angle
with the vertical, determine the tension in the string.

Could you please double check my solution? I think the solution is

F= mg + mgh F = (2.2) (9.8) + (2.2) (9.8) (0.36) F 29.32

or

F = mg +0.5 m (V^2) = (2.2) (9.8) + 0.5 (2.2) ( 2.62 ^2) = 29.32 what do you think ?

Answer 1

Date 7100 11 '100 Page SET NUTTOCOK Aus. ☆ 4= e-l coso e lcoso h = e(1-coso) Th TH e = 2.8m v=4.7ml mg ©= ago mg Energy conservations mght Imva = I mv² 2 Vo = 2 v2-29h Ev -age (1-010) 2 vo - (4.712 - 2x9.8x28 (1-COS 89") foi Vo = 3.8999 m Is velocity at given position Tension (T) = mut ma coso e T= (212)*(3.899912 2.8 + Qia)x19.8 % Cos 29 Thugi Tension in string (I) = 30.807 Ao. Please upute the answer your method À Lorong.