DESPERATE Physics help?

Question

i have 3 problems i cant solve due tomorrow and are graded...Please include how to get to the answer

1.A 5.0kg is moving 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall. After the block collides with the spring the spring is compressed a maximum distance of .68m. What is the speed of the block when the spring is compressed to only one-half the max distance


2.A block of mass M is moving along the horizontal frictionless surface with a speed of 5.7 m/s. If the slope is 11(degrees) and teh coefficiant of kinetic friction .26 how far does the block travel up the incline

3A block of 2.0kg is placed on a compressed vertical spring that is compressed .050m the spring is released and it propels the block vertically upward. when the block has risen .60m above its initial postion its velocity is 1.7m. how much potential energy was stored in the spring


Thanks feel free to answer one or all i'll take all the help i can get...

Answer 1

1.A 5.0kg is moving 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall. After the block collides with the spring the spring is compressed a maximum distance of .68m. What is the speed of the block when the spring is compressed to only one-half[1/2?] the max distance

Initial energy = 0.5 x m x v^2


let say maximum distance = x

so compressed = (1/2)x

therefore
Energy got into spring = 0.5 k. (0.5 x)^2
= 0.5^3. k. x^2
[
you need to find "k"
so
k.x^2 = m.v^2
x= 0.68

k= (m.v^2)/(x)^2
]
= 0.5^3. k. x^2

= (0.5^3)(x^2)((m.v^2)/(x)^2)
= (0.5^3)(m.v^2)

therefore energy left to the that moving thing

= 0.5(mv^2)-(0.5^3)(m.v^2)

New kinetic enery = 0.5.m.V^2

V^2 = (v^2)- (0.5^2)(v^2)

v=5

put that in there

so new velocity = 4.33


and dunn trust that my answer is right>


2.A block of mass M is moving along the horizontal frictionless surface with a speed of 5.7 m/s. If the slope is 11(degrees) and teh coefficiant of kinetic friction .26 how far does the block travel up the incline

Initial energy = 0.5.m.v^2

Once it starts on the slope..
Gravity works against it and friction works against it

find work for both gravity and friction..

Gravity
=F(g).d d is the distance it travelled
F(g) = mg.sin (A) A = 11.1 or whatever

so work by gravity = mg.sin (A). d

work by friction = f. mg.sin (A). d f= coeefecitn of friction

so
mg.sin (A). d(1+f) = 0.5.m.v^2

d= (0.5 v^2)/(g. sin (A). (1+f))
=6.89 m


3A block of 2.0kg is placed on a compressed vertical spring that is compressed .050m the spring is released and it propels the block vertically upward. when the block has risen .60m above its initial postion its velocity is 1.7m. how much potential energy was stored in the spring

So, Total energy when the block is high up in the air
= 0.5.m.v^2 + mgh
h= 0.6 and v=1.7

and then initially.. that block that no energy
so potential in the spring = that above thing
=14.7 J

ain't so sure about this one... it can't be so easy!!!!!!!!

Answer 2

I do not take physics... But I am close to definite that the minimal coefficient of kinetic vigour might be got by way of making use of the distinct equal drive towards the item (10kg weight). The pace that's gift after a 3m drop over the distance of 5m is: three/five = zero.60. Therefore the coefficient, or factorial significance of minimal kinetic vigour required to give up the item is strictly zero.60 (in concept). @Additional main points: Oops! yeah I was once very fallacious. I concept it was once might be a less complicated reply.

Answer 3

They all sound like Newton's three laws of motion problems. You might be better off calling a friend who can help and explain what equations apply where.

Answer 4

1. one half the distance, one quarter of the work
2. work, work, work
3. potential energy plus work

Answer 5

Do you actually expect ANYONE to read, let alone, answer, your long winded homework for you?