A body of mass 50 kg falling from a certain height is
brought to rest after striking the ground with a speed of 5
m/s. If the resistance force of the ground is 500 N find
the duration of the contract

Respuesta :

Answer:

t = 0.5 seconds

Explanation:

Given that,

Mass of a body, m = 50 kg

Initial speed, u = 0

Final speed, v = 5 m/s

The resistance force of the ground = 500 N

We need to find the time of contact. The formula for the force acting on an object is given by :

F = ma

a is acceleration

[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{50\times (5-0)}{500}\\\\t=0.5\ s[/tex]

So, the time of contact is 0.5 seconds.

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