A 2.1 kg pendulum bob hangs from a 2 meter string. It is pulled to the side, position A, so that the bob is 125 cm (h on the sketch) above the lowest position, position C. The bob is released and begins to swing. How fast is it moving at that point?
mgh=1/2 mv^2
v=sqrt (2gh) h is 1.25m
To determine the speed of the pendulum bob at position C, we need to find the gravitational potential energy at position A, and then use the principle of conservation of energy.
The gravitational potential energy (PE) of an object at height h above a reference point is given by the equation:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
In this case, the mass of the pendulum bob is 2.1 kg and the height h is 125 cm (converted to 1.25 meters). The acceleration due to gravity is approximately 9.8 m/s^2.
PE_A = 2.1 kg * 9.8 m/s^2 * 1.25 m
= 25.725 J
Now, at position C, all of the potential energy has been converted into kinetic energy (KE). The equation for kinetic energy is given by:
KE = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
Using the principle of conservation of energy, we equate the potential energy at position A to the kinetic energy at position C:
PE_A = KE_C
25.725 J = (1/2) * 2.1 kg * v^2
Simplifying the equation:
50.85 J = 2.1 kg * v^2
Divide both sides by 2.1 kg:
v^2 = 50.85 J / 2.1 kg
v^2 = 24.214 m^2/s^2
Take the square root of both sides to find v:
v = sqrt(24.214 m^2/s^2)
v ≈ 4.92 m/s
Therefore, the pendulum bob is moving at a speed of approximately 4.92 m/s at position C.