Question

A grubby toddler views a box of chocolate on a 2 meter high countertop. The box will fall with the energy of 14 Joules.
a)What is the work done?
b)What is the Force of the box of chocolates when it lands? Givens: Formula: Sub: Solve:

Answer 1

Interpretation note: the box falling from 2 m gains 14 J of kinetic energy before it lands. “Work done” by gravity equals that energy. The impact force depends on how quickly (or over what distance) the box is brought to rest — that information is not given, so I give the general formula and some example values.

a) Work done by gravity
Givens:
- h = 2.0 m
- E = m g h = 14 J (given)
- g = 9.8 m/s²

Formula:
W = ΔPE = m g h

Sub:
W = 14 J (given)

Solve:
Work done by gravity = 14 J.

You can also find the mass:
m = E / (g h) = 14 / (9.8 × 2) = 14 / 19.6 = 0.714 kg (≈0.71 kg).

b) Force when it lands
If by “force when it lands” you mean the weight (force of gravity) on the box:
Givens:
- m = 0.714 kg
- g = 9.8 m/s²

Formula:
F_weight = m g

Sub:
F_weight = 0.714 × 9.8

Solve:
F_weight ≈ 7.0 N.

If you mean the average impact force on landing (F_avg), you need a stopping distance d (or stopping time t). Use:
F_avg = (kinetic energy) / d = E / d
or equivalently F_avg = m v² / (2 d), with v = √(2 g h) ≈ 6.26 m/s.

Example values (using E = 14 J):
- If d = 0.01 m (1 cm): F_avg = 14 / 0.01 = 1400 N
- If d = 0.05 m (5 cm): F_avg = 14 / 0.05 = 280 N
- If d = 0.10 m (10 cm): F_avg = 14 / 0.10 = 140 N

Summary:
a) Work done by gravity = 14 J.
b) Weight = 7.0 N. Average impact force depends on stopping distance; F_avg = E/d (examples above).