Answer:
a) 4.65m/s
b) 59.8 N , 1.01125 N
Explanation:
a)
m = mass of the ball = 1 kg
r = length of the string = 2.0 m
h = height gained by the ball as it moves from lowest to topmost position = 2r = 2 x 2 = 4 m
v = speed at the lowest position = 10 m/s
v' = speed at the topmost position = ?
Using conservation of energy
Kinetic energy at topmost position + Potential energy at topmost position = Kinetic energy at lowest position
(0.5) m v'² + m g h = (0.5) m v²
(0.5) v'² + g h = (0.5) v²
(0.5) v'² + (9.8 x 4) = (0.5) (10)²
v' = 4.65m/s
b)
T' = Tension force in the string when the ball is at topmost position
T = Tension force in the string when the ball is at lowest position
At the topmost position:
force equation is given as
[tex]mg + T' = \frac{m v'^{2}}{r}[/tex]
[tex](1)(9.8) + T' = \frac{(1) (4.65)^{2}}{2}[/tex]
T' = 1.01125 N
At the lowest position:
force equation is given as
[tex]T - mg = \frac{m v^{2}}{r}[/tex]
[tex]T - (1) (9.8) = \frac{(1) (10)^{2}}{2}[/tex]
T = 59.8 N
