A2.00 kg pendulum bob on a string 1.50 m long is released with a velocity of 0.00 m/s when the support string makes an angle of 60.0 degrees with the vertical. what is the tension in the string at the bottom of the swing

39.2 Newtons First, calculate how high the bob is above the bottom of the swing. Since the string is making a 60 degree angle from the vertical, the angle the string makes against the horizon is 30 degrees. And the sine of 30 degrees is 0.5. So the vertical distance from the top of the pendulum is 0.5 * 1.5 m = 0.75m and therefore the distance the bob will drop is 1.5m - 0.75m = 0.75m. Now at the point of release we have all gravitational potential energy and that will be: 2 kg * 9.8 m/s^2 * 0.75 m = 14.7 kg*m^2/s^2 = 14.7 Joules At the bottom of the swing all of that potential energy will have been converted to kinetic energy. The equation for that is E = 0.5 MV^2 14.7 J = 0.5 * 2.0 kg * V^2 14.7 kg*m^2/s^2 = 1 kg * V^2 14.7 m^2/s^2 = V^2 3.834057903 m/s = V Now the tension in the string at the bottom of the swing will be the sum of the gravitational attraction and the centripetal force. The equation for centripetal force is: F = MV^2/r Where F = Force M = Mass V = velocity r = radius We already know that V^2 is 14.7 m^2/s^2 from calculating the velocity above, so we have F = M*14.7 m^2/s^2/1.5 m F = 2.0 kg * 14.7 m^2/s^2 / 1.5 m F = 29.4 kg*m^2/s^2 / 1.5 m F = 19.6 kg*m/s^2 And the gravitational attraction is F = 2 kg * 9.8 m/s^2 = 19.6 kg*m/s^2 So the total force experienced is 19.6 kg*m/s^2 + 19.6 kg*m/s^2 = 39.2 kg*m/s^2 = 39.2 N So the tension in the sting at the bottom of the swing is 39.2 newtons.

Wish i knew this to !