What Is the Maximum Acceleration of a 75g Object on a Spring?

In summary, a 75g object with a maximum displacement of 0.080m to the right vibrates at the end of a spring with a constant of 5.0 N/M. Using the equation Fs=-kx, we can calculate that the maximum acceleration is approximately -5.3ms/^2 to the left. The correct SI unit for mass is kilograms, not grams.
  • #1
Imbadatphysics
7
0

Homework Statement


A 75g object vibrates at the end of a spring(k=5.0 N/M) if the maximum displacement of the object is 0/080m to tge right what is the maximum acceleration?

Homework Equations

The Attempt at a Solution


Fs=-kx so... (-5)(0.080)=Fs and fs=fg so -0.4=75a then a is -.0053... the answer is like 5.3ms/^2 left did i convert something wrong...
 
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  • #2
Note the unit for the mass. What is the SI unit for mass?
 
  • #3
grams.
 
  • #4
The SI unit of mass is not grams.
 
  • #5
oh. you use kg.. i ssee thanks
 
  • #6
OK. By the way, welcome to PF!
 
  • #7
thanks ;D
 

What is Hooke's Law?

Hooke's Law is a physical principle that describes the relationship between the force applied to an object and the resulting extension or compression of the object.

Who discovered Hooke's Law?

Hooke's Law was discovered by English scientist Robert Hooke in the 17th century.

What is the formula for Hooke's Law?

The formula for Hooke's Law is F = -kx, where F is the force applied, k is the spring constant, and x is the displacement of the object from its equilibrium position.

What are the limitations of Hooke's Law?

Hooke's Law is only applicable to elastic materials, meaning materials that can return to their original shape after being deformed. It also assumes that the force applied is directly proportional to the displacement of the object, which may not always be true.

Why is Hooke's Law important?

Hooke's Law is important because it helps us understand and predict the behavior of elastic materials under different forces. It is also the basis for many engineering applications, such as designing springs and shock absorbers.

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