Thalia climbs on a swing which is at 0.593 m above ground. Her father pulls back

Question

Thalia climbs on a swing which is at 0.593 m above ground. Her father pulls back the swing to a height of 1.34 m above ground and pushes so that its initial speed is 3.35 m/s. The combined mass of Thalia and swing is 30.8 kg. At the lowest point of the motion, Thalia reaches down and grabs her knapsack of mass 10 kg that is lying on the ground. Note: Neglect air resistance and assume that the swing is frictionless.

1) Find how high above the ground Thalia, swing and knapsack will rise. (meters)

2) After reaching maximum height, Thalia swings backwards and when she reaches the lowest point of the motion, she simply lets go of her knapsack that drops to the ground. Find the maximum height that Thalia and swing reach on the backward swing.

Explanation / Answer

First we find the velocity of Thalia at the moment before when she picks up her knapsack. That is

0.5*m*v02 + m*g*(h1-h0) = 0.5*m*v12, where v0= 3.35 is the initial velocity, h1 = 1.34 the heigh from the ground, h0= 0.593 the height of the swing to the ground, and v1 is the velocity when she reaches the lowest point.

v1 = 5.08m/s

Now we write the energy equilibrium formula from the moment she grabs the knapsack to when she reaches the highest point moving forward.

0.5*m2*v12 = m2*g*(h2 - h0), where h2 is the final height going forward. We notice that the height doesn't depend on the mass of thalia and her knapsack

h2 = 1.91 m

2)

from the first part we know that the velocity before thalia loses her knapsack will be 5.08m/s. To get the new height she will reach, we again use energy:

m*g*(h3 - h0) = 0.5*m*5.082

h3 = 1.91 m

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