Question

A 0.5 kg box drops from a height of 10 m above the ground without it spinning as it falls. Assuming that air resistance is negligible, when the box is 5 m above the ground,

(a) What is its kinetic energy?

(b) What is its linear speed?

Answer 1

according to the conservation of energy the total mechanical energy is always remains conserved.

Mechanical energy = kinetic energy + potential energy

→at height of 10m,

Mechanical energy = (1/2)*m*v² + m*g*h

using the values,

Mechanical energy = 0 + (0.5)*9.81*10

Mechanical energy = 49.5 Joules

so total mechanical energy of the box is 49.5 Joules

(a)→ at height of 5 m,

Mechanical energy = Kinetic energy + potential energy

as mechanical energy remain always conserved.

49.5 = K.E. + (0.5)*9.81*5

24.75 J = K.E

Thus the kinetic energy at height of 5 m is 24.75 Joules.

(b) as kinetic energy is expressed as,

K.E. = (1/2)*m*V²

as we know kinetic energy at 5 m height is 25.75 J.

24.75 = 0.5 * 0.5 * V²

solving for V,

V = √98

V = 7√2 m/s

V = 9.89 m/s

Thus the linear speed of box is 9.89 m/s.