What is the speed of the 40 kg box after it has gone up a ramp that's 3 meters tall? The box is initially going at V = 20 meters/second.

Would the equation just be 1/2 mv = 1/2 mv? Or do you incorporate gravitational energy which is = mgh?

Correct V = 18.43 m/s

Explanation:

Let be initial velocity (speed) V₀ and the current speed at some point V

At start, the box had kinetic energy which is equal to the total energy.

Ek₁ = E = m v₀²/2 = 40 · 20²/2 = 8,000 J

At the ramp point 3m high the box had potential energy:

Ep = m · g · h = 40 · 10 · 3 = 1,200 J  we took that g= 10 m/s²

The total energy at any time is equal :

E = Ek + Ep ⇒ Ek = E - Ep = 8,000 - 1,200 = 6,800 J

Ek = m V²/2 ⇒ V = √2 · Ek/m = √ 2 · 6,800/40 = √340 = 18.43 m/s

V = 18.43 m/s

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c. magnitude and direction