Question

a simple pendulum consists of a 2.0kg mass attached to a string.it is relesed from rest at X as shown. the speed at the lowest point Y is

Answer 1

Answer:6m/sec

Explanation:h= 1.85

1/2mv^2= mgh

v^2= 2*g*h

v=(2*9. 8*1. 85)^1/2

=6

Answer 2

Answer:

v = 6.02 m/s

Explanation:

It is given that,

Mass of the object, m = 2 kg

Let X is the point that lies in the upward side and Y is the point that lie in the down side. At point Y, the potential energy is equal to the kinetic energy. Using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

Let h = 1.85 m

$v=\sqrt{2\times 9.8\times 1.85}$

v = 6.02 m/s

So, the speed at the lowest point is 6.02 m/s. Hence, this is the required solution.