Unit conversion difficulties on simple problem

AI Thread Summary
To determine the work needed to change the velocity of a 55 g particle from 27 m/s to 41 m/s, the correct approach involves using the Work-Energy theorem, which relates work to the change in kinetic energy. The initial calculations mistakenly focused on momentum rather than energy. Converting grams to kilograms is necessary for consistency in units, as work is measured in joules. The correct formula for work in this context is W = ΔKE = 0.5 * m * (v_final^2 - v_initial^2). Understanding the relationship between kinetic energy and work is crucial for solving the problem accurately.
xgoddess210
Messages
27
Reaction score
0

Homework Statement



A 55 g particle is moving to the left at 27 m/s. How much work must be done on the particle to cause it to move to the right at 41 m/s? Answer in joules

Homework Equations





The Attempt at a Solution



27-(-41)=68 , 68*55= 3750

the units here are grams m/s. I need joules. I have no clue on how to make that happen. I can change the g to kg and even get it to Newtons (since I know a joule= Newton-meter) , but I can't take it any farther. Please help!
 
Physics news on Phys.org
xgoddess210 said:
27-(-41)=68 , 68*55= 3750

the units here are grams m/s. I need joules. I have no clue on how to make that happen. I can change the g to kg and even get it to Newtons (since I know a joule= Newton-meter) , but I can't take it any farther. Please help!

Just convert g to kg in your equation is all you need to do.
 
I tried that- It gives me an answer of 3.74, but that is wrong. This doesn't really make sense to me.
 
Once you have the correct units, your calculation is nonetheless in error. You have W = m\Delta v , which is the change in momentum, not the work done. What is the correct equation to use for Work?
 
oh! The equation I have for work is W=Fx, but all that is given to me is two velocities and a mass. Force is m*a so w=m*a*x. I guess I'm still at a loss here.
 
Oops. That's right. I just thought you were converting units wrong like your title was asking.

Overlooked that it was work and that's not work you have.

What about figuring the change in kinetic energy then?
 
Are you at all familiar with the Work-Energy equation or Conservation of Total Energy equations??
 
Back
Top