Calculating Velocity of a Block Launched by an Ideal Spring

[king_naeem]

An ideal spring is used to fire a 15.0-g block horizontally across a frictionless table top. The spring has a spring constant of 20 N/m and is initially compressed by 7.0 cm. The speed of the block as it leaves the spring is:


i do the following:

energy stored by spring: (kv^2)/2= 0.049

when the spring is at equilibrium position all the energy wwould be in kinetic thus i solve for the velocity,

but i get the wrong answer...what am i doing wrong?

 

[ZapperZ]

An ideal spring is used to fire a 15.0-g block horizontally across a frictionless table top. The spring has a spring constant of 20 N/m and is initially compressed by 7.0 cm. The speed of the block as it leaves the spring is:


i do the following:

energy stored by spring: (kv^2)/2= 0.049

when the spring is at equilibrium position all the energy wwould be in kinetic thus i solve for the velocity,

but i get the wrong answer...what am i doing wrong?

Er... if you meant by "v" above as "velocity", then you need to double check the expression for the potential energy of a compressed spring. Hint: it involves the compressed LENGTH, and not dependent on any velocity.

Zz.

 

[HallsofIvy]

It's hard to tell what you are doing wrong since you don't show us HOW you solve for velocity!

The potential energy stored in the spring, with spring constant 20 N/m, by compressing it 0.07 meters is (1/2)(20)(0.07)²= 0.049 Joules.

Once the spring has "uncompressed", all of its potential energy goes into the kinetic energy of the block so (1/2)(0.015kg)v²= 0.049 and v= 2.56 m/s approximately.

 

[king_naeem]

aah...my problem was that i forgot to convert 15 grams into kilograms...ahh!