Average force exerted on the coin by the soil

[terainfizik]

Homework Statement

A coin of mass 10g is dropped from the top of a tall building. If after failing for 10s , it falls intosoil and penetrates to a depth of 5cm, calculate

(Q) the average force exerted on the coin by the soil.

Homework Equations

The Attempt at a Solution

Should I use F=ma ? or use F= m( v-u ) / t ? to get the solution ?
Give me hint !

 

[vanesch]

Homework Statement

A coin of mass 10g is dropped from the top of a tall building. If after failing for 10s , it falls intosoil and penetrates to a depth of 5cm, calculate

(Q) the average force exerted on the coin by the soil.

Homework Equations

The Attempt at a Solution

Should I use F=ma ? or use F= m( v-u ) / t ? to get the solution ?
Give me hint !

Given that they ask for the average force and that anyway you don't know anything else but the begin and end velocities (in the soil), it is the second of your formulas that you should use.

 

[terainfizik]

Given that they ask for the average force and that anyway you don't know anything else but the begin and end velocities (in the soil), it is the second of your formulas that you should use.

The end of the velocities ( in the soil ) should be zero ? Is that mean F= m(v-u)/t , where m= weight of coin , v= ending speed of dropping into the soil ? and u=initial speed of dropping from the building ( zero also ? )
Guide me please.

 

[vanesch]

The end of the velocities ( in the soil ) should be zero ? Is that mean F= m(v-u)/t , where m= weight of coin , v= ending speed of dropping into the soil ? and u=initial speed of dropping from the building ( zero also ? )
Guide me please.

The end velocity in the soil is zero of course, because when it has penetrated its 5 cm, I guess it is not moving anymore. The begin velocity in the soil is however, the velocity the object obtained in free fall.

So you should first solve the free fall problem to get this "final" velocity of the fall (which is the initial velocity of the next stage of motion: the penetration in the soil).

 

[Doc Al]

Should I use F=ma ? or use F= m( v-u ) / t ? to get the solution ?

The first equation would work just fine, once you calculate the acceleration. Since you don't have the time that it takes the coin to come to rest, that second equation won't help (unless you figure out the time first!).

Another approach is to consider the work done on the coin by the soil. That's the approach I would recommend.

 

[vanesch]

The first equation would work just fine, once you calculate the acceleration. Since you don't have the time that it takes the coin to come to rest, that second equation won't help (unless you figure out the time first!).

Another approach is to consider the work done on the coin by the soil. That's the approach I would recommend.

That's also an idea.

The way I wanted to propose this was:
you calculate the average force by having "start velocity minus final velocity"/t, but with t as of yet unknown.

That average force gives you (containing the unknown t) the average acceleration, and then you use s = a(t)/2 t^2 = 5cm to solve for the unknown t.

 

[Doc Al]

It's all good--any of these three approaches will work just fine, if you are careful. So pick one and get busy!

Note: Once the coin hits the soil there are two forces acting on it.

 

[ShawnD]

Another approach is to consider the work done on the coin by the soil. That's the approach I would recommend.

I think Doc is right on this one. I can't seem to solve this problem using momentum.
0 = (m)(v) + (F1)(t1) - (F2)(t1); t1 is the impact time, F1 is gravity, F2 is soil
0 = (m)(g)(t2) + (m)(g)(t1) + (F2)(t1); t2 is fall time (10s)
0 = (0.010)(9.8)(10) + (0.010)(9.8)(t1) + (F2)(t1)
F2 is the answer to the problem and t1 is unknown
t1 is derived from:
d = (1/2)(a)(t^2); d is 5cm, a is unknown
a is derived from:
(v1 - v2)/t; where t is the same t as the above equation

It's like one would need to substitute a bunch of things over and over again to get the right answer.

Energy is very straight forward
0 = (1/2)(m)(v^2) + (m)(g)(d) - (F)(d); d is the soil distance of 5cm, F is soil
0 = (1/2)(m)[(g)(t)]^2 + (m)(g)(d) + (F)(d)
0 = (1/2)(0.010)[(9.8)(10)]^2 + (0.010)(9.8)(0.05) - (F)(0.05)
Solve for F

 

[terainfizik]

Thank you for giving me these hints . I have worked it out . You guys shows clear statements working .

 

[vanesch]

I think Doc is right on this one. I can't seem to solve this problem using momentum.
0 = (m)(v) + (F1)(t1) - (F2)(t1); t1 is the impact time, F1 is gravity, F2 is soil
0 = (m)(g)(t2) + (m)(g)(t1) + (F2)(t1); t2 is fall time (10s)
0 = (0.010)(9.8)(10) + (0.010)(9.8)(t1) + (F2)(t1)
F2 is the answer to the problem and t1 is unknown
t1 is derived from:
d = (1/2)(a)(t^2); d is 5cm, a is unknown
a is derived from:
(v1 - v2)/t; where t is the same t as the above equation

Yes, but this is easy:

v2 = 0
a = v1/t, hence d = 1/2 v1/t (t^2) = 1/2 v1 t from which follows t = 2 d / v1
F = m v1/t = m v1^2/(2 d)

But if you want to take into account the extra m.g acting, then we have to add this: F' = F + m.g

Now, it is true that the energetic approach is clearly the most straightforward one...