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.
1 answer
W = ΔKE =KE2 –KE1 = m•v²/2 – 0 = m•v²/2 =50•25/2 =625 J
4 answers