Question

a 10 kg rock begins to fall off a 10 m building. what is the potential kinetic, total energies and velocities at 10 m,7.5 m,5m, 2.50 m, and at ground level

Answer 1

This can be done using the concept ofof conservation of energy

That is total energy of the rock stays constant always.

At 10 m height the velocity of rock is 0.

The total energy = PE + KE

PE = mgh, where m is mass=10 kg, g is acceleration due to gravity = 9.8 m/s² and h is height = 10 m.

KE = 1/2 .( mv² ) , v is the velocity.

Potential energy at 10m = mgh = 10*9.8*10= 980Joules

The kinetic energy at 10 m = 1/2 m v²

But v= 0 so

KE = 1/2 *10*0² = 0 J

Total energy = 980 + 0 = 980 J

At height 7.5 m the total energy = 980 J

Potential energy = mgh = 10*9.8*7.5= 735 J

The kinetic energy = total energy- potential energy

KE = 980- 735 = 245 J

Ke = 1/2 m v² .= 245

So the velocity at the point 7.5 m ,

v² =245*2/m = 245*2/10= 49 so v= 7

Or velocity at 7.5m = 7 m/s

The total energy at 5 m= 980 J

The potential energy at 5 m = mgh = 10*9.8*5= 490 J

The kinetic energy at 5 m = total energy-potential energy= 980- 490 = 490 J

KE = 1/2mv² = 490J

V² = 2*490/m = 2*490/10= 9.8 so v= 9.9 m/s

So velocity at 5m = 9.9m/s

The total energy at 2.5 m = 980J

Potential energy at 2.5m = mgh = 10*9.8*2.5= 245 J

Kinetic energy at 2.5 m = 980- 245 =735 J

KE = 1/2 mv² = 735J

V² = 3*735/m = 2*735/10 = 147 so v = 12.12m/s

So the velocity at 2.5m = 12.12 m/s

At ground level height h = 0

Total energy = 980 J

So potential energy= mgh = 10*9.8*0 = 0J

Kinetic energy= total - potential energy= 980-0

Kinetic Energy at ground = 980 J

KE = 980 = 1/2 mv²

V² = 2*980/10 = 196 or v= 14 m/s

So the velocity at ground level = 14 m/s

(Please upvote if helpful)