Answer:
3.9096m/s
Explanation:
Since the ball moved in circular path, then it will experience centripetal acceleration which can be expressed below
a= v^2/r.......eqn(1)
From second law of Newton, we know that the ball will experience both force of tension and that of gravity.
F= (T - Fg)...........eqn(2)
r= distance= 1.5m
Where T= force of tension= 10N
Fg= force of gravity
But we know that
F= ma...............eqn(3)
Where a= acceleration
m= mass of the object= 500g= 0.5kg
Substitute eqn(1) and (2) into 3
(T - Fg)= mv^2/r
v^2= (T - Fg)r/m
v^2= √(T - Fg)r/m
But Fg= mg
v^2= √[10 -( 0.5×9.81)])1.5/0.5
v=3.9096m/s
Hence, the speed of the ball at that point is 3.9096m/s
Answer:
3.91m/s
Explanation:
The mass of the ball(m)=500g = 0.5kg
The radius of the string is(r)=1.5m
The tension in the string is(T)=10N
The acceleration due to gravity = [tex]9.8m/s^{2}[/tex]
The tension in the string when the body is at the bottom is given by
[tex]T = \frac{mv^{2} }{r}+mg[/tex]
To find the speed of the ball, we make v the subject of the formula
Therefore, [tex]v=\sqrt\frac{r(T-mg)}{m}[/tex]
[tex]v=\sqrt\frac{1.5(10-0.5*9.8)}{0.5}[/tex]
[tex]v=\sqrt\frac{7.65}{0.5}[/tex]
[tex]v=\sqrt15.3=3.9115[/tex] ≈ 3.91m/s
Therefore, the speed of the ball is 3.91m/s
