What is the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s ?
Please explain
2007-10-06 16:13:34 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
I worked the same problem a month ago, gave me hell. From Physics Vol. I by Giancoli, right?
use v^2 = vo^2 + 2a(Δx) where vo = original velocity
13^2 = (0)^2 + 2a(2.8)
a = 30.179 m/s^2
and since F = ma, F = (30.179 m/s^2)(7 kg) = 211.25 N
2007-10-06 16:25:27 · answer #1 · answered by swash2314 3 · 0⤊ 0⤋
Work = F*d
KE = (1/2)*m*v^2
I will assume the shot comes to rest such that W = KE1 - KE2, where KE2 = 0
KE = 0.5*7*13^2 = 591.5
591.5 = Favg*2.8
Favg = 591.5/2.8 = 211.25
Please verifty this as it has been a long time since I've done this.
2007-10-06 16:21:15 · answer #2 · answered by J S 2 · 1⤊ 0⤋
energy is conserved.
How much energy work is transfer to kinetic energy
work = force * distance = Fx
KE = .5 * mass * speed^2 = .5mv^2
Fx = .5mv^2
F(2.8) = .5(7)(13)^2
F = 211.25N
you can solve this problem by using New's 2nd law
F = ma
find acceleration
Vf^2 = 2ad + Vi^2
13^2 = 2(a)(2.8)
a = 30.178 m/s^2
F = 30.178 * 7
F =~ 211.25N
2007-10-06 16:26:08 · answer #3 · answered by 7 · 0⤊ 0⤋