Question

6) Consider a 10.0 kg block of wood that's released from rest at a distance 20.0m above the

ground and assume ground level is the point of zero gravitational potential energy. Calculate:

a) the initial gravitational potential energy of the block upon release

b) the translational kinetic energy of the block just before it hits the ground.

c) the translational kinetic energy of the block when it has fallen half the distance to the
ground.

d) the speed of the block when it has fallen half the distance to the ground.

Answer 1

Step-by-step explanation:

a) Potential energy is weight times height.

PE = mgh

PE = (10.0 kg) (9.8 m/s²) (20.0 m)

PE = 1960 J

b) Energy is conserved. As the block falls, potential energy is converted to kinetic energy. Just before the block lands, all of the potential energy has been converted to kinetic energy.

KE = 1960 J

c) When the block has fallen half the distance, half the potential energy has been converted to kinetic energy.

KE = 1960 J / 2

KE = 980. J

d) Kinetic energy equals half the mass times the square of the velocity.

KE = ½ mv²

980. J = ½ (10.0 kg) v²

v = 14.0 m/s