A 10 kg block moving at 10 m/s in a direction 45 degrees above the horizontal. When it has fallen to a point that is 10 m below the initial point measured vertically (without air friction), the block's kinetic energy is closest to

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Answer:

Kinetic energy = 1500 J

Explanation:

The computation of the block's kinetic energy is shown below:

As we know that

Conservation of energy is

PE_i + KE_i = PE_f + KE_f

where,

Initial Potential energy = PE_i = m gh = 10kg× 10m/s^2 × 10m = 1000 J

Initial Kinetic energy = KE_i = (0.5) m V^2 = (0.5) (10 kg) (10 m/s)^2 = 500 J

Final potential energy = PE_f = mgh = 0      

As h = 0 which is at reference line

So

PE_i + KE_i = PE_f + KE_f

Now put these valeus to the above formulas

1000 J + 500 J = 0 + KE_f

After solving this

Kinetic energy = 1500 J

The block's kinetic energy is closest to 1500 Joules.

Kinetic energy :

The energy is always conserved.

So that, the total kinetic energy will be sum of initial potential energy and kinetic energy during falling.

Given that, mass(m)=10kg, v=10m/s, h=10m,g=10m/s^2

              K.E=(1/2)mv^2 + mgh

              K.E=(1/2)*10*100 + (10*10*10)

              K.E=500 + 1000=1500Joule

The  block's kinetic energy is closest to 1500 Joules.

Learn more about the kinetic energy here:

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