Ima Bratt is at the park with her father. The 20.0 kg Ima is on a swing following the path as shown below. Ima has a speed of 0 m/s at position A and is a height of 3.0 m above the ground.

Question

At position B, Ima is 1.0 m above the ground. At position C (2.0 m above the ground), Ima ejects from the swing seat and travels as a projectile along the path shown. At point F, Ima is a negligible distance above the ground. Assume negligible resistance throughout the motion. How much work must Ima's father do to raise her up to point A? How much KE does Ima have point B?How much speed does Ima have point F? NEED ANSWER QUICK!!

Answer 1

(a) To bring her to an elevation of 3.0m above the ground the dad has to do negative work.

Wa = -mgha = -20 * 9.81 * 3 = -588.6 N*m (J)

(b) To find kinetic energy at B

Apply Work Energy Principle

A - > B

Sum of Work = Change in Kinetic Energy

A to B generates positive work, force of weight in the direction of motion

+ m * g * (3-1) = (1/2) * m * (V_B² - V_A²₎

20 * 9.81 * 2 = (1/2) * 20*V_B²

Solve for Speed at B = 6.26 m/s

KE_B = (1/2) * m * V_B² = 392.4 J

(c) Speed at F

Sum of Work = Change in Kinetic Energy

A to F generates positive work, force of weight in the direction of motion

+ m * g * (3) = (1/2) * m * (V_F² - V_A²₎

+ m * g * (3) = (1/2) * m * (V_F²)

V_F = 7.67 m/s