Please see the attached file.

____ 1. A simple pendulum, 2.0 m in length, is released with a push when the support string is at an angle of 250 from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, what is its speed at the bottom of the swing? (g = 9.8 m/s2)
a. 2.3 m/s
b. 2.6 m/s
c. 2.0 m/s
d. 1.8 m/s
e. 0.5 m/s
Total Energy at release point = 0.5*mv02 + mg(h - hcos250)
Total Energy at the bottommost point = 0.5*mv2
Now, 0.5*mv2 = 0.5*mv02 + mg(h - hcos250)
Or, v2 = v02 + 2gh(1-cos250) = 1.22 + 2*9.81*2*(1-cos250) = 5.116499
Therefore, v =

____ 2. A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the speed received in the above situation?
a. 1.41 d
b. 1.73 d
c. 2.00 d
d. 4.00 d
e. 5.35 d
From energy balance, ½kx2 = ½m(2d)2
Therefore, x =