Question

using potential and kinetic energy quick check
nans suitcase is on a bench. the suitcase has a mass of 14 kilograms. nan accidentally knocks
the suitcase off the bench, and the suitcase falls at a speed of 2.7 meters per second. how much
kinetic energy does the suitcase have just before it hits the floor?
ke = \frac { 1 } { 2 } m v ^ { 2 }
(1 point)
18.9 j
102.06 j
204.12 j
51.03 j

Explanation:

Step1: Substitute values into formula

Given \(m = 14\space kg\), \(v=2.7\space m/s\), and \(KE=\frac{1}{2}mv^{2}\).
Substitute: \(KE=\frac{1}{2}\times14\times(2.7)^{2}\)

Step2: Calculate \((2.7)^{2}\)

\((2.7)^{2}=2.7\times2.7 = 7.29\)
So \(KE=\frac{1}{2}\times14\times7.29\)

Step3: Calculate \(\frac{1}{2}\times14\)

\(\frac{1}{2}\times14 = 7\)
Now \(KE=7\times7.29\)

Step4: Multiply

\(7\times7.29 = 51.03\)

Answer:

51.03 J