Asked by Chris
An applied force accelerates a 50-kg crate along a frictionless floor from rest to 5 m/s.The work performed by this force is most nearly
125 J
250 J(x)
500 J
625 J
1,250 J
Can someone please explain this to me and help? I thought it was 250 J but my calculation was wrong.
125 J
250 J(x)
500 J
625 J
1,250 J
Can someone please explain this to me and help? I thought it was 250 J but my calculation was wrong.
Answers
Answered by
Elena
W = ΔKE =KE2 –KE1 = m•v²/2 – 0 = m•v²/2 =50•25/2 =625 J
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.