Monique has a mass of 46 kg and she is sitting on a swing that hangs 0.50 m above the ground. Pierre pulls the swing back and releases it when the seat is 1.40 m above the ground. How fast is Monique moving when the swing passes through its lowest point?

The answer is 4.2m/s, but I don't know how to arrive at the answer.

The Potential Energy at the start of the swing has to equal the KE at the bottom.

mg(1.40-.50)=1/2 m v^2

v=sqrt((2*.90)g)=4.2 m/s

To find the speed at which Monique is moving when the swing passes through its lowest point, we can use the principle of conservation of energy.

1. The initial potential energy of Monique at the highest point (1.40 m above the ground) is given by:
PE_initial = m * g * h_initial
where m is Monique's mass (46 kg), g is the acceleration due to gravity (9.8 m/s^2), and h_initial is the height of the swing at the highest point (1.40 m).

2. The final potential energy of Monique at the lowest point (0.50 m above the ground) is given by:
PE_final = m * g * h_final
where h_final is the height of the swing at the lowest point (0.50 m).

3. Since energy is conserved, the initial potential energy is equal to the final kinetic energy:
PE_initial = KE_final

4. The kinetic energy at the lowest point is given by:
KE_final = (1/2) * m * v^2
where v is the velocity or speed of Monique at the lowest point.

5. Equating the potential energy at the highest point to the kinetic energy at the lowest point, we can solve for the velocity:
m * g * h_initial = (1/2) * m * v^2
Since mass m appears on both sides of the equation, it cancels out.

6. Simplifying the equation, we get:
g * h_initial = (1/2) * v^2

7. Plugging in the given values:
g = 9.8 m/s^2
h_initial = 1.40 m

9.8 m/s^2 * 1.40 m = (1/2) * v^2

8. Solving for v:
v^2 = 2 * 9.8 m/s^2 * 1.40 m

v^2 = 27.44 m^2/s^2

9. Taking the square root of both sides, we find:
v = √(27.44 m^2/s^2)

v ≈ 5.24 m/s

Therefore, Monique is moving at a speed of approximately 5.24 m/s when the swing passes through its lowest point.