Question

A 2.5 kg banana falls from a bunch that is 3.5 m above the ground. Calculate the speed with which the banana will strike the ground.

If the banana makes a tiny 0.050 m imprint in the mud, what force must the ground exert on the banana?

Answer 1

1) According to the conservation of energy law, the initial potential energy of the banana

$$W= mgh$$

is equal to its final kinetic energy

$$K = \dfrac{mv^2}{2}$$

Here $m = 2.5kg$ is the mass, $g = 9.8m/s$ is the gravitational acceleration, $h = 3.5m$ is the initial height, $v$ is the speed with which the banana will strike the ground. Expressing $v$, obtain:

$$mgh = \dfrac{mv^2}{2}\\ v = \sqrt{2gh}\\ v = \sqrt{2\cdot 3.5\cdot 9.8} \approx 8.28 m/s$$

2) According to the work-energy theorem, the change in the kinetic energy of the banana is equal to the work experienced by it from the mud. The kinetic energy has changed from

$$K_1 = \dfrac{mv^2}{2}$$

to $K_2$ . In turn, the work is defined as follows:

$$A = F_{net}s$$

where $F_{net }$ is the force the ground exerted on the banana, and $s = 0.05m$ is the distance travelled by the banana in the mud. Thus, obtain:

$$\dfrac{mv^2}{2} = F_{net}s\\ F_{net} = \dfrac{mv^2}{2s} = \dfrac{2mgh}{2s} = \dfrac{mgh}{s}\\ F_{net} =\dfrac{2.5\cdot 9.8\cdot 3.5}{0.05} = 1715N$$

Answer. 1) 8.28 m/s, 2) 1715 N.