How long it takes for the ice block to hit the ground

[chawki]

Homework Statement

5kg of ice falls away from the edge of the roof of a block of flats, at a height of 26m above the ground.

Homework Equations

ignoring the air drag, find out:
a) how long it takes for the ice block to hit the ground.
b) what is the speed at which the ice block hits the ground.
c)how much energy it transfers to the surroundings when it comes to a stop.

The Attempt at a Solution

a)
y=1/2*g*t²
t=\sqrt{}y/0.5*g
t=\sqrt{}26/0.5*9.8
t=2.3s and that's the time it takes for the ice block to hit the ground.
b)
V=g*t
V=9.8*2.3
V=22.54 m/s
c)
E=1/2*m*V²
E=1/2*5*22.54²
E=1270.129 joules

 

[Aureum]

Looks good but I would start with energy as it is more accurate because you don't lose any figures.

U=h*m*g or Gravitational Potential Energy=(Height)(Mass)(Gravity)

1274 joules= 26*5*9.8

 

[chawki]

Looks good but I would start with energy as it is more accurate because you don't lose any figures.

U=h*m*g or Gravitational Potential Energy=(Height)(Mass)(Gravity)

1274 joules= 26*5*9.8

There is no potential enegry at h=0
It's the kinetic energy.

 

[Peter G.]

What Aureum means is that you used figures you got as answers to find the kinetic energy at the bottom.

The kinetic energy at the bottom is equal to the Gravitational Potential Energy at the top, as he showed: 26.5*5*9.8

 

[chawki]

Yes but we hae been asked to find the energy at ground.

 

[Doc Al]

Yes but we hae been asked to find the energy at ground.

But mechanical energy is conserved.

 

[chawki]

Yes and that's Ke+Pe
Pe=0
so only Ke is left. right Doc Al ?

 

[Doc Al]

Yes and that's Ke+Pe
Pe=0
so only Ke is left. right Doc Al ?

Of course only KE is left. But that KE equals the original PE. That's the best way to calculate it with the least amount of round-off error.

 

[chawki]

What is the correct writting?
Mechanical energy at ground = Ke+Pe (Pe at ground =0)
so Mechanical energy at ground =Ke = 1/2*5*22.54^2 = 1270.129 ?

 

[Doc Al]

What is the correct writting?
Mechanical energy at ground = Ke+Pe (Pe at ground =0)
so Mechanical energy at ground =Ke = 1/2*5*22.54^2 = 1270.129 ?

Do you agree that KE(at ground) = PE(at roof)?
If you agree, calculate the PE(at roof) = mgh and compare.

What you did is not "wrong", but unnecessary and introduces round-off errors since you are using a rounded-off value for the velocity. (That was Aureum's point.) What if they asked question c first?

(Of course, quoting an answer to 7 significant figures is a bit silly, since your data is only good to 2 significant figures. Nonetheless...)

 

[chawki]

I agree with KE(at ground) = PE(at roof) but what is the name of the law that we should mention for that?
we can say: by applying the conservation of mechanical energy we have KE(at ground) = PE(at roof)
difference of mechanical energy between ground and roof = 0 ?

 

[Doc Al]

Yes, it's just conservation of mechanical energy.

Note that we've been assuming that you've covered that topic in your course. If not, then the way you solved it was perfectly fine.