45. 3. (III) A 7.26 kg bowling ball hangs from the end of a 2.5 m rope. The ball is pulled back until the rope makes an angle of 45 with the vertical. a. What is the increase in the ball’s potential energy when it is lifted? b. If the ball is released from rest and swings downward like a pendulum, how much kinetic energy will the ball have at the bottom of its swing? c. How fast will the ball be moving at the bottom of its swing?

Explanation:

For the first question look at the picture. It should be pretty clear. Remember that cos45° = sqrt2 / 2.

2) We calculate it thanks to the conservation of mechanical energy:

Em1 = Em2

U1 + K1 = U2 + K2

K1 = 0J

U2 = 0j

U1 = K2

3) mgh = 1/2mv^2

v = sqrt(2gh)

no. dot product   =0 when vecs at right angles

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.