the string of pendulum is 2.0 m long,the bosy is pulled sideway so that the string becomes horizontal and the bob is released what is the speed with which the bob is released ?what is the speed with which the bob arrives the lowest piont?
To calculate the speed at which the bob is released and the speed at the lowest point, we can use the principles of conservation of energy.
Let's assume that the bob is released from a point where the string is horizontal. At this point, all the potential energy is converted to kinetic energy.
1. Calculate the potential energy at the starting point:
The potential energy (PE) of an object at a certain height is given by the equation PE = m * g * h, where m is the mass of the bob, g is the acceleration due to gravity (approximated as 9.8 m/s²), and h is the height.
Since the bob is released from the horizontal position, the height (h) is equal to the length of the string (2.0 m).
PE = m * g * h
2. Calculate the kinetic energy at the starting point:
The kinetic energy (KE) of an object is given by the equation KE = (1/2) * m * v², where m is the mass of the bob and v is the velocity.
At the starting point, the velocity (v) is equal to the speed at which the bob is released (let's call it v1), which is what we're trying to find.
KE = (1/2) * m * v1²
3. Set the potential energy equal to the kinetic energy:
Since energy is conserved, we can equate the potential energy to the kinetic energy:
PE = KE
m * g * h = (1/2) * m * v1²
4. Calculate v1 (the speed at which the bob is released):
To find v1, we cancel out the mass (m) and solve for v1:
g * h = (1/2) * v1²
v1 = sqrt(2 * g * h)
Substituting the values, we get:
v1 = sqrt(2 * 9.8 * 2.0)
v1 = sqrt(39.2)
v1 ≈ 6.26 m/s
Therefore, the speed at which the bob is released is approximately 6.26 m/s.
5. Calculate the speed at the lowest point (when the bob is at the bottom of its swing):
At the highest point (where the bob was released), all the potential energy is converted to kinetic energy. At the lowest point, all the kinetic energy is converted back to potential energy.
Since energy is conserved, the potential energy at the lowest point is equal to the kinetic energy at the starting point:
PE = KE
m * g * 0 = (1/2) * m * v_lowest²
Since the potential energy at the lowest point (PE) is zero, we can solve for v_lowest:
0 = (1/2) * v_lowest²
v_lowest = 0
Therefore, at the lowest point, the bob comes to a momentary halt, and its speed is 0 m/s.