BECE Science 2025 Paper

Question : 8

Total: 40

A mango of mass 2 kg hangs 15 m on top of a mango tree. Determine the value of the kinetic energy attained just before it hits the ground. [

g = 10 \text{ m s}^{-2}

]

30.0 J
225.0 J
300.0 J
3.0 J

Validate

Solution:

Concept:

As the mango falls, its potential energy is changed into kinetic energy (air resistance is ignored). Just before it hits the ground, all the potential energy it had at the top has become kinetic energy.

Formula used:

E_p = mgh

and

E_k

(just before impact)

= E_p

(at the top)

Given:

m = 2 \text{ kg}

h = 15 \text{ m}

g = 10 \text{ m s}^{-2}

Solution:

E_p = mgh = 2 \times 10 \times 15 = 300 \text{ J}

By conservation of energy,

E_k = E_p = 300.0 \text{ J}

Key point:

30 \text{ J}

comes from forgetting

m

(

mgh

without the 2 gives 150, without

g

gives 30). Always multiply all three:

m \times g \times h

Answer:

Option C (300.0 J)

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