Question

A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its kinetic energy at circled A? 1.2635 Correct: Your answer is correct. J (b) What is its speed at circled B? 4.54 Correct: Your answer is correct. m/s (c) What is the net work done on the particle by external forces as it moves from circled A to circled B?

Answer 1

a).
`E_(kA)=1.2635 J`

b).
`V_(B)=4.535(m)/(s)`

c). Δ
`E_(t)=8.4635 J`

Step-by-step explanation:

ΔE=kinetic energy

a).

`E_(kA)=(1)/(2)*m*v_(A) ^(2) \\ v_(A)=1.9 (m)/(s)\\ m=0.70kg\\E_(kA)=(1)/(2)*0.70kg*(1.9 (m)/(s))^(2) \\E_(kA)=1.2635 J`

b).

`E_(kB)=(1)/(2)*m*v_(B) ^(2)`

`V_(B)^(2)=(E_(kB)*2)/(m) \\V_(B)=\sqrt{(E_(kB)*2)/(m)} \\V_(B)=\sqrt{(7.2J*2)/(0.70kg)} \\V_(B)=4.53 (m)/(s)`

c).

net work= EkA+EkB

`E_(t)=E_(kA)+ E_(kB)\\E_(t)=1.2635J+7.2J\\E_(t)=8.4635J`