A 200kg gorilla sits on a spring with a spring constant of 5000N/m. The gravitational field strength is 9.8N/kg.?

a) How much shorter will the spring become when it is sat on?

b) How much GPE is lost by the gorilla when the spring compresses?

c) How much energy will be stored in the spring while the gorilla is sitting on the spring?

d) Compare the answers to parts b and c and explain their relative sizes.

1 Answer

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  • oubaas
    Lv 7
    4 weeks ago

    a) How much shorter (x) will the spring become when it is sat on?

    deflection x = m*g(k = 9.8*200/5000 = 19.6/50 = 0.392 m

    b) How much GPE is lost by the gorilla when the spring compresses?

    lost GPE = m*g*x = 200*9.8*0.392 = 768.3 joule

    c) How much energy SE will be stored in the spring while the gorilla is sitting on the spring?

    SE = m*g*x/2 = k*x*x/2 = k/2*x^2 = 2500*0.392^2 = 384.15 joule

    d) Compare the answers to parts b and c and explain their relative sizes.

    SE = GPE/2; if the spring force was constant throughout the whole deflection , then the gorilla would never compress the spring

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