× Didn't find what you were looking for? Ask a question

Top Posters
Since Sunday
17
6
6
5
s
5
t
5
s
5
9
5
g
5
h
5
l
5
o
5

# Potential and Kinetic Energy Problem-high school physics?

wrote...
Posts: 82
Rep:
10 years ago
 Potential and Kinetic Energy Problem-high school physics? A constant force of 15.3 N acts upward on a iron block that weighs 11.7 N. (a) If the block is initially at rest, what is its kinetic energy 6.53 s after the force is applied? What is the increase in the potential energy of the body 6.53 s after the force is applied. Read 701 times 1 Reply

### Related Topics

Replies
wrote...
10 years ago
 Given:Fa = 15.3 N upwardFg = 11.7 N downwardFind:a)  K.E. = ?  at t = 6.53 sec and  Vi = 0b)  delta P.E. = increase in potential energySolution:a)  K.E. = (1/2)mv^2     K.E. = (1/2)(Fg/g)v^2     K.E. = (1/2)(11.7/9.8)v^2Solving for v, final velocity:     Fnet = ma     Fa - Fg = (Fg/g)(a)     a = ((Fa - Fg)/(Fg/g)     a = (15.3 - 11.7)/(11.7/9.8)     a = 4.3 m/s^2     a = (v - vi)/t     v = vi + at     v = 0 + (4.3m/s^2)6.53s     v = 28.1 m/sHence,K.E. = (1/2)(11.7/9.8)v^2K.E. = (1/2)(11.7/9.8)(28.1)^2K.E. = 471 joules   ANS.b)  delta P.E.g = mgh     delta P.E.g = (11.7)(h)Solving for h:         2ah = v^2 - vi^2         h = (v^2 - vi^2)/2a         h = (28.1^2 - 0)/2(4.3)         h = 91.8 m      delta P.E.g = (11.7)(91.8)      delta P.E.g = 1074.8 joulesteddyboy
Explore
277 People Browsing
264 Signed Up Today
Related Images

234

200

359