PLEASE PLEASE PLEASE HELP me to do these Energy PHYSICS questions, Thanks :)?

Question

PLEASE PLEASE PLEASE HELP me to do these PHYSICS questions, Thanks :)


1. a) If the KE of an arrow is quadrupled, by what factor has its speed increased?

b) If the speed is tripled, by what factor does its KE increase?

2. How much work must be done to stop a 1070 kg car traveling at 111 km/h?

3. One car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 5.5 m/s, they then have the same kinetic energy. What were the original speeds of the two cars?

first car=.....m/s
second car=.....m/s

Don't worry about # 1 question a & b, i already know how to do it, i wasn't sure but it was right

however, please help on #2 & #3, PLEASE
tHANKS

Answer 1

KE=.5*m*v^2

a) KE2=4*KE1

KE1=.5*m*v1^2

KE2=.5*m*v2^2

since KE2=4*KE1
4*KE1=.5*m*v2^2
or
KE1=.5*m*v2^2/4
and
.5*m*v1^2=.5*m*v2^2/4
simplify
v1^2=v2^2/4
take the sqrt of both sides
v2=2*v1

b) similar
KE1=.5*m*v1^2

KE2=.5*m*v2^2
v2=3*v1
so
KE2=.5*m*9*v1^2

KE2/KE1=9

2. The work is equal to the kinetic energy

3)
m1=2*m2

KE1i=KE2i/2
or
2*.5*m1*v1i^2=.5*m2*v2i^2
using mass relationship

2*.5*2*m2*v1i^2=.5*m2*v2i^2
simplify
4*v1i^2=v2i^2
also

KE1o=KE2o
where
.5*m1*(v1i+5.5)^2=
.5*m2*(v2i+5.5)^2

using mass relationship
2*.5*m2*(v1i+5.5)^2=
.5*m2*(v2i+5.5)^2

simplify
2*(v1i+5.5)^2=(v2i+5.5)^2

now we have two equations and two unknowns
2*(v1i+5.5)^2=(v2i+5.5)^2
4*v1i^2=v2i^2
or
2*v1i=v2i
and
sqrt(2)*(v1i+5.5)=v2i+5.5
and
sqrt(2)*(v1i+5.5)=2*v1i+5.5

v1i=5.5*(sqrt(2)-1)/
(2-sqrt(2))

j


j

Answer 2

The vertical top is 1m while the block is on the incredible of the airplane the flexibility would be thoroughly capacity. E = mgh= 10 kg x 10 m/s^2 x one million m= one hundred J Gravitational stress is conservative. artwork completed via it (the flexibility won via the container does no longer count on the path taken between the preliminary and the suitable positions.

Answer 3

1) KE = 1/2m*v^2
the mass of an arrow is constant. so KE = const*v^2

2) Work energy theorem (energy conserved)

KE1 +PE1 +W = KE2 +PE2

no change in height, so PE1 = PE2 = 0

At stop, KE2 = 0

so you have KE1 = -W

KE = 1/2*m*v^2
m = 1070kg
v = 111km/hr*(1000m/km)*(1hr/3600s)

Plug n chug

3) I got lost, you're on your own

Answer 4

1.

a) 2

KE = 1/2mv^2

(1/2) is constant
(m) is constant
KE up by (x4)
v^2 up by (x4)
==>v up by (x square root of 4)
==> v up by (x2)

b) KE = 1/2mv^2

Same type of thing. v up by (x3) ==> KE up by x(3^2) ==> KE up by (x9)

Note: I used (x2) for times 2, etc.

Boss walked in. Back in a few.

Edit:
Ok, I think odu83 is working on it. I'll leave this up, though.