a catcher "gives" with the ball when he catches a 0.113 kg baseball moving at 21.9 m/s. If he moves his glove a distance of 7.24 cm, what is the average force acting on his hand? Answer in units of kN.
What if his glove and hand move 10.1 cm? Answer in units of N.
2006-10-07 11:33:22 · 6 answers · asked by Dee 4 in Science & Mathematics ➔ Physics
Still can't figure this out, any other suggestions????????????
2006-10-07 11:51:20 · update #1
Force x Distance moved is work done or loss in energy.
In the given case the loss in energy is the kinetic energy of the ball which is
0.5 m v v = 0.5 x 0.113 x 21.9 x 21.9 = 27.1 joule.
Therefore the force is “energy / distance” = 27.1 / 0.0724 =374 newton in the first case and 27.1 / 0.101 =268 newton in the second case.
2006-10-07 13:57:24 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
The second answer is on the right track. The ball's energy is 0.5*m*v^2. You are given m and v. This energy is dissipated by (and equal to) the force times the distance and you are given the distance (in this case, two distances). Everything is in compatible units so this should be a snap; the force you calculate will be in N. Just remember to scale for cm to m in both cases and N to kN in the first case.
2006-10-07 19:03:30 · answer #2 · answered by kirchwey 7 · 1⤊ 0⤋
apply relation of distance, time, acc.
v^2 - u^2 =2as
where s = 7.24cm
final velocity (v) = 0
initial velocity(U)=21.9m/s
put in equation and find the value of acc.
u know F=M*A put value of a which u had already found out
u r knowing m =.113
evaluate and find the answer
for second part repeat the procdure
2006-10-07 22:36:49 · answer #3 · answered by n nitant 3 · 0⤊ 0⤋
Cant be boovered to work it out but i think u need to use a Force of motion as the ball is accelerating at a constant speed. you may need to do sum transposition to get the rite equation.
2006-10-07 19:30:59 · answer #4 · answered by Mark 2 · 0⤊ 1⤋
taking a guess on this one ... to lazy to look up what i long ago forgot.
i believe work done is equal to the change in energy (maybe i just made that up)
work done is also equal to force times distance (oh man been so long since ive done this stuff)
energy is kinetic and is (m*v^2)/2
should be enough to do the problem
2006-10-07 18:44:01 · answer #5 · answered by Anonymous · 1⤊ 1⤋
most stupid question ever
2006-10-07 18:42:01 · answer #6 · answered by JOHN jen 4 · 0⤊ 4⤋