Using Energy Conservation to solve this problem: Jumping on a spring scale

[r12214001]

Homework Statement:

my cal:760(1+X)=0.5kX²

Relevant Equations:

760(1+X)=0.5kX²

I suppose spring compression to be X when jumpping from 1m. Therefore gravational potential mgh=760(1+X) and my cal:760(1+X)=0.5kX²
why the solution manual state that 760=0.5kX²+760X

 

[jbriggs444]

I suppose spring compression to be X when jumpping from 1m. Therefore gravational potential mgh=760(1+X) and my cal:760(1+X)=0.5kX²
why the solution manual state that 760=0.5kX²-760X

Algebra.

Start with$$760(1+X)=0.5kX^2$$ and subtract $760X$ from both sides to get $$760(1+X)-760X=0.5kX^2-760X$$Now on the left hand side notice that you can cancel that 760X and get $$760=0.5kX^2-760X$$

 

[r12214001]

Algebra.

Start with$$760(1+X)=0.5kX^2$$ and subtract $760X$ from both sides to get $$760(1+X)-760X=0.5kX^2-760X$$Now on the left hand side notice that you can cancel that 760X and get $$760=0.5kX^2-760X$$

typing error
solution manual state that 760=0.5kX²+760X

 

[PeroK]

typing error
solution manual state that 760=0.5kX²+760X

The solution manual took $X$ to be negative.

 

[PeroK]

PS my bathroom scales only go up to about $150kg$ or $1,500N$. I'm not able to test this out.

 

[r12214001]

PS my bathroom scales only go up to about $150kg$ or $1,500N$. I'm not able to test this out.

OK i didnt notice that. thanks for correction