A70 kg person does a bungee jump from a bridge. the natural length of the bungee cord is 20m and it has a spring constant k of 80 n/m. when the bungee cord has a total length of 25m, what is the acceleration of the jumper (to 2 significant figures)? hint: draw a fbd showing the magnitudes and directions of all forces on the jumper at the instant described.

Explanation:

For this exercise we can use Newton's second law and ask us for the acceleration of the jumper

    F = ma

Let's set a reference system and take the positive direction up

    fe - W = m a

    k DX - mg = m a

we clear we calculate

     a = k / m ( -x₀) - g

     a = 80/70 (25 -20) - 9.8

     a = 11.43 -9.8

     a = 1.6 m / s²

Directed up

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

less than one you can't create energy