a man is diving off a 3m spring board. If the man has a mass of 60kg and compresses the spring board of k = 15,000N/m by 0.4m, with what velocity does the man enter the water?

Question

please respond using this formulas

gpe=mgh

ke=1/2 mv^2

f=kx

epe=1/2 k x^2

work =fs

power= w/t

Answer 1

At the moment he launches from the board, his kinetic energy (KE) equals the potential energy of the compressed spring (kx²/2). In addition, he has the potential energy from an initial height of h₀. I suggest solving this problem using energy conservation.

m = mass of man

h₀ = height of board

x = compression length of spring

k = spring constant

Moment of launch:

E = mgh₀ + KE = mgh₀ + kx²/2

Moment of reaching water:

E = mg(0) + mv²/2 = mv²/2

By energy conservation,

mv²/2 = mgh₀ + kx²/2

v = √[(2/m) (mgh₀ + kx²/2)]

You know m, g (9.8 m/s²), h₀, k, and x. Plug in the given numbers and get v in m/s.