Answer:
(a) [tex]a=6.883m/s^{2}[/tex]
(b) [tex]T_{tension}=0.24N[/tex]
Explanation:
Given data
Angle α=35.0°
Mass m=20g=0.02 kg
To find
(a) Acceleration a
(b) Tension T
Solution
For part (b) tension T
From Newtons Second Law the Tension on string is given b[tex]T_{tension} Cos(\alpha )=W_{weigth}\\T_{tension}Cos(\alpha )=m_{mass}g_{gravity}\\T_{tension}=\frac{m_{mass}g_{gravity}}{Cos(\alpha )}\\T_{tension}=\frac{0.02kg*9.8m/s^{2} }{Cos(35^{o} )}\\T_{tension}=0.24N[/tex]
For part(a) acceleration
The acceleration is given by:
[tex]F=ma=TSin(\alpha )\\ma=TSin(\alpha )\\a=\frac{TSin(\alpha )}{m}\\ a=\frac{(0.24N)Sin(35 )}{0.02kg}\\a=6.883m/s^{2}[/tex]
