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Kelli weighs 435 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back and releases it when the seat is 1.00 m above the ground. Assume that air resistance is negligible.A) How fast is Kelli moving when the swing passes through its lowest position? B) If Kelli moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction?

User Finswimmer
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2 Answers

3 votes

Final answer:

Kelli's speed when the swing passes through its lowest position is 4.14 m/s. The work done on the swing by friction is 905 N.

Step-by-step explanation:

To determine the speed of Kelli when the swing passes through its lowest position, we can use the principle of conservation of mechanical energy. At the highest point, Kelli has potential energy and no kinetic energy. At the lowest point, the potential energy is zero and the entire mechanical energy is in the form of kinetic energy. Therefore, the speed of Kelli at the lowest position can be found using the equation:

Kinetic energy = Mechanical energy - Potential energy

We know that the mechanical energy is the same at both points, so we can write:
0.5 mv^2 = mg
Where m is the mass of Kelli, v is her speed, g is the acceleration due to gravity, and h is the difference in height between the highest and lowest points of the swing. Rearranging the equation, we have:
v = sqrt(2gh)

Substituting the given values, we get:
v = sqrt(2 * 9.8 m/s^2 * 1.36 m) = 4.14 m/s

Therefore, Kelli is moving at a speed of 4.14 m/s when the swing passes through its lowest position.

To find the work done on the swing by friction, we can use the work-energy principle. The work done by friction can be calculated using the equation:
Work = Change in kinetic energy

Given that the initial velocity of the swing is 2.0 m/s and the final velocity is 0 m/s due to friction, the change in kinetic energy is equal to the initial kinetic energy. Therefore, the work done on the swing by friction is:
Work = 0.5 * m * (initial velocity)^2 = 0.5 * 435 N * (2.0 m/s)^2 = 905 N

User RickAndMSFT
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4.4k points
4 votes

Answer:A 3.57m/s

B Workdone= 246.44J

Explanation:Given: change in heigt= 1-0.36=0.64, mass=435/9.8= 44.4Kg

PE=mgh=44.4×9.8×0.64=278.4J

Let PE=KE=1/2mv2

278.4=44.4/2×v^2

278.4=22.2v^2

V^2=278.4/22.4

V^2=12.79

V=3.57m/s

B. Velocity at the bottom of the swing is 2m/s

KE=44.4×0.36×2=-31.96J

Workdone=PE+KE=278.4-31.96=246.44J

User Dshukertjr
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4.3k points