A 1 kg flashlight is dropped from rest and falls to the floor without air resistance . At the point during its fall, when it is 0.7 meters above the floor, its potential energy exactly equals its kinetic energy. How fast is the flashlight moving at this point, in m/s 3.7

The speed of the flashlight at that point is 3.7 m/s

Explanation:

When an object of mass M is at a height H above the ground, the potential energy of the object is:

U = M*H*g

Where g is the gravitational acceleration, g = 9.8 m/s^2

And for an object with velocity v, the kinetic energy is:

K = (M/2)*v^2

We know that when the flashlight of mass  1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:

M = 1kg

H = 0.7m

g = 9.8 m/s^2

Replacing these in the equations, we get:

U = K

(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2

As the mass factor appears in both sides, we can remove it:

(0.7 m)*(9.8 m/s^2) = (v^2)/2

Now we can multiply both sides by 2:

2*(0.7 m)*(9.8 m/s^2) = v^2

Now let's apply the square root to both sides:

√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s

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