Question

Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back and releases it when the seat is 1.09 m above the ground. The acceleration of gravity is 9.8 m/s 2 . How fast is Betty moving when the swing passes through its lowest position? Answer in units of m/s.

Answer 1

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = $\frac{1}{2} *m*vf^{2}$  (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

$vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s$

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.