Question

PLEASE HELP

A 2.80 kg mass is dropped from a
height of 4.50 m. Find its kinetic
energy (KE) when it is 3.00 m
above the ground.

Answer 1

Answer: 42 m/s

Step-by-step explanation:

m=2.8 kg , V1=0 m/s , h1=4.5 m , h2=3 m , V2=? m/s

E1 = E2 >>> K1 + U1 = K2 + U2

Kinetic Energy (KE) = K =
`(1)/(2)` m
`V^(2)` Potential Energy (PE) = U = mgh

`(1)/(2)` m
`V1^(2)` + mg
`h_(1)` = K2 + mg
`h_(2)` >>> V1=0 >> K1=0 >>> K2 = -mgΔh

K2 = 2.8 kg × 10 m/s2 × 1.5 m = 42 m/s

Answer 2

41.16J

Step-by-step explanation:

m =2.8 kg

h1 = 4.5 m

h2 = 3 m

PE at top = mgh1= 4.5mg

PE at 3 m = mgh2= 3mg

energy is conserved los in PE = gain in KE

KE at 3 m above the ground

= mg(4.5 - 3) J

= 2.8 kg*9.8m/s^2*1.5 m

= 41.16 joules