Calculating Change in Kinetic Energy of 0.23kg Object

[brad sue]

Hi I have a problem here and I am a little bit confused.

An object of mass 0.23kg is initially at tyhe origin and is acted on by the sole force F=(0.50 N)i. After a certain time, the object is at a position r=(.88m)i.
What is the change in the object's kinetic?

I tried to solve this problem by th equatiom W_net=K, but I do not get the answer of the textbook. They found .89m/s .Apparently,it v. They did not specify.

Please can youhelp

Thank you for your time
B

 

[HallsofIvy]

Sorry, this makes no sense to me! If the object moves .88 m under a .5 N force, then it has had .44 J of work done on it. It's kinetic energy should have changed by .44 J. If, as you say, that was the sole force then we should have (1/2)(.23 kg) v²= 0.44. that gives v²= 0.88/.23= 3.8 so v= 1.9 m/s.

 

[brad sue]

Sorry, this makes no sense to me! If the object moves .88 m under a .5 N force, then it has had .44 J of work done on it. It's kinetic energy should have changed by .44 J. If, as you say, that was the sole force then we should have (1/2)(.23 kg) v²= 0.44. that gives v²= 0.88/.23= 3.8 so v= 1.9 m/s.

Yes, I found the same result as you.
I must have a typo in my textbook.
Thank you anyway

B