An 0.80-kilogram object tied to the end of a 2.0-meter string swings as a pendulum. at the lowest point of its swing, the object has a kinetic energy of 10.0 joules. what is the speed of the object at the instant that the string makes an angle of 50.0° with the vertical?

3.3 m/s

Explanation:

As the object rises above the lowest point, some of the kinetic energy is converted to potential energy.  From the diagram, we can see that at angle θ, the object rises to height h:

h = L - L cos θ

Conservation of energy:

KE₀ = KE + PE

KE₀ = 1/2 mv² + mgh

Substituting:

KE₀ = 1/2 mv² + mg(L - L cos θ)

Given KE₀ = 10.0 J, m = 0.80 kg, g = 9.8 m/s², L = 2.0 m, and θ = 50.0°:

10.0 = 1/2 (0.80) v² + (0.80) (9.8) (2.0 - 2.0 cos 50.0)

v = 3.32 m/s

Rounding to 2 sig-figs, the speed of the object is 3.3 m/s.

the total distance traveled by the car is

30km + 49km = 79 km .

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