Find Speed of Block After Released from Compressed Spring

[jk2455]

Homework Statement

A 1.5kg block is held against a spring of force constant 1.5×10⁴ N/m, compressing it a distance of 0.20m .How fast is the block moving after it is released and the spring pushes it away?(m/s)

answer=20m/s
i am not sure what to do but i tried this and obviously didnt get the right answer could you please see what i am doing wrong and explain it to me

Homework Equations

F=-kx
$$\Sigma$$F=ma

v²=v₀²+2a x

Homework Statement

given:
m=1.5kg,k=1.5×10⁴ N/m,x=0.20m,v₀=0

unknown:
time(s), acceleration(m/s^s), velocity(m/s)

2. The attempt at a solution

F =ma=-kx
a=-kx/m
a=-3000

v²=v₀²+2ax
v=$$\sqrt{}$$2ax<---(that is supposed to say the square root of 2ax i don't know why it looks like that)
v=34.6m/s

 

[hotvette]

The potential energy of the compressed spring gets converted to kinetic energy of the mass upon release.

 

[PhanthomJay]

Homework Statement

A 1.5kg block is held against a spring of force constant 1.5×10⁴ N/m, compressing it a distance of 0.20m .How fast is the block moving after it is released and the spring pushes it away?(m/s)

answer=20m/s
i am not sure what to do but i tried this and obviously didnt get the right answer could you please see what i am doing wrong and explain it to me

Homework Equations

F=-kx
$$\Sigma$$F=ma

v²=v₀²+2a x

Homework Statement

given:
m=1.5kg,k=1.5×10⁴ N/m,x=0.20m,v₀=0

unknown:
time(s), acceleration(m/s^s), velocity(m/s)

2. The attempt at a solution

F =ma=-kx
a=-kx/m
a=-3000

If this equation was correct, which it is not, then your math would imply that a = -2000. Both solutions are incorrect, including the use of the minus sign.

v²=v₀²+2ax
v=$$\sqrt{}$$2ax<---(that is supposed to say the square root of 2ax i don't know why it looks like that)
v=34.6m/s

You can do it this way, but you need the correct value of the acceleration. Since the force of a spring is not constant (it varies with x) , then the acceleration is not constant either, as it too varies with x. You can use average force and average acceleration in lieu of the calculus to solve the problem, but it is much easier to use energy methods, if you are familiar with them.

 

[jk2455]

thanks a lot both of you(hotvette &PhanthomJay) i got the correct answer =)

 

[jk2455]

one question
is x always positive even if the spring is stretched or compressed??

 

[PhanthomJay]

one question
is x always positive even if the spring is stretched or compressed??

No, it can be either. The negative sign in F= -kx comes from the fact that the force of the object on the spring is opposite to the direction of motion. In your problem, assuming the object is on the right end of the spring and moving to the right, then the force on the object acts in the direction of the accelerataion, to the right, thefore positive net force implies positive acceleration.