Sure hope someone can help with this one.....?

Question

A 537 g block starts at rest on a slope 4.7 m up (the highest point) and slides down a frictionless track. It leaves the track at 2.2 m high (where the track becomes horizontal), & then flies thru the air & subseqently strikes the ground. Acceleration of gravity is 9.81 m/s^2.

1. What is the speed (v) of the block when it leave the track, in units of m/s?

2. What is the horizontal distance (x) from when the block leaves the track, travels thru the air & hits the ground of the block, in units of m?

3. What is the speed of the block when it hits the ground, in units of m/s?

Answer 1

1. Entire energy in the object should be a constant. For a frictionless track, in this situation, ony the weight applies on the object. So entire energy at the beginning should be equal to the entire energy at the second instant.

mgh1=mgh2+1/2(mv^2)
gh1 = gh2 + (v^2)/2
v^2 = 2g(h1-h2)

Therefore,
v = 7 m/s

2. After the block leave the track, only gravitational force is applied on it. It will travel 2.5m (4.7-2.2) to hit the ground. When it leave the track, the vertical velocity is zero. Horizontal velocity is 7m/s.

Apply S=ut+(at^2)/2 Vertically.

S = 0 + (at^2)/2
t^2 = 2S/a
t = 0.714 s

Apply S=ut+(at^2)/2 Horizontally. (here a = 0)

S = 7*0.714 m
S = 5 m

3. Apply v = u + at Vertically

When the object left the track, vertical velocity is zero. a = 9.81

So v = at
v = 9.81*0.714
v = 7 m/s

Answer 2

1. Use conservation of energy: mgh = 1/2 mv^2 where h is the height above which it leaves the track.

2. First find the time it takes to hit the ground using the equation of motion in the y-direction: dy = -1/2 gt^2.
Second, use this time in the equation of motion in the x-direction to get delta x: dx = vt where v is the answer in (1).

3. The square of the speed will be the sum of the squares of the x and y components of velocity. The x component won't change from the answer in (1). The y-component is given by v = -gt where t was found in (2).

Good Luck

Answer 3

1. KEi+PEi = KEf+PEf
KEi=0
PEf-PEi=mg(H1-H2)=(.537 kg)*(9.8 m/s^2)*(4.7-2.2 m)=13.157 J
KEf=13.157 J = mv^2
v = Sqrt(13.157 J / .537 kg) = 4.95 m/s

2. there are no horizontal forces on the block after it leaves the track, so compute the time it is in the air and multiply by the velocity:

(Xf-Xi)=(1/2)g*t^2
t=Sqrt(2*2.2 m/9.8 m/s^2) = .67 s

d = v*t = 4.95 m/s*.67 s = 3.32 m

3. to find the vertical speed, consider the block is in freefall for 2.2 meters.

V = Sqrt(2*g*Xf-Xi) = Sqrt(2*9.8 m/s^2*2.2 m) = 6.57 m/s

horizontal speed does not change while it is on the air, so that is still 4.95 m/s. To get the total speed, use pythagorean theorem V= Sqrt(6.57^2+4.95^2) = 8.22 m/s

Answer 4

GPE + KE=0 In other words, all the gravitational potential energy is converted to kinetic energy (no loss to friction)

so, m x g x h + 1/2mv^2=0

plug the numbers in(don't forget to use the proper units!!!!!), solve for v. 9.6 m/s I think.
For question 2, you know the initial velocity from above, and since horizontal velocity is CONSTANT, you need to know how long it took to fall from 2.2 m d=ViT + 1/2AT^2
Vi is zero, so 2.2m=1/2(9.8)T^2 get T
velocity/time=distance
For #3, Vf^2=Vi^@ +2A(2.2m)

GOOD luck!

Answer 5

potential energy at top = mgh1. poential energy at 2.2 = mgh2.
difference in potential energy = kinetic energy= 1/2 mv^2. so we get v^2=5g.
now this is constant horizontally. vertical velocity when it hits ground is u^2 = 2as = 4.4g. so sqrt v^2 + u^2 gives magnitude of velocity. now we can find tan also and get direction.
to find the distance. 2.2 = 0.5 g t^2. now we know t. multiply it to v to get the distance.

Answer 6

Angle of the slope or total length of the slope is required to find the answer.

Answer 7

Looks like someone forgot to do their homework.