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The question is: A body of mass 10 g falls from a height of 3 m into a pile of sand. The body penetrates the sand at distance of 3 cm before stopping. What force has the sand exerted on the body?

The solution is: Let $v$ be the velocity of the body at the instant it reaches the pile of sand Then from the relation $v=v_0^2+2 g y$, we have $$ \begin{aligned} v^{2} &=0+2 \times\left(9.8 \text {m/s}^2\right) \times 3 \,\text{m:} \\ &=58.8\,(\text{m/s})^{2} \end{aligned} $$ This velocity is reduced to zero due to the deceleration ' $a$ ' produced by the sand. Thus, from the relation $u^{2}=v_{0}^{2}+2 a y$, we have $$ \begin{array}{l} 0=58.8+2 a(0.03 \text {m}) \\ a=-\frac{58.8}{2 \times 0.03}=-980\,\text {m/s}^{2} \end{array} $$ The mass of the body is $10 \mathrm{g}=0.01\, \mathrm{kg}$. Hence the (retarding) force exerted by the sand on it is $$ \begin{aligned} F &=m a \\ &=0.01\,\mathrm{kg} \times\left(-980\,\mathrm{m} / \mathrm{s}^2\right) \\ &=-9.8\,\mathrm{N} \end{aligned} $$

Now My question is if the answer is correct cause when the body will reach on the surface the gravity will still work. So, do I need to add 9.8 m/s^2 in the retardation.

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  • $\begingroup$ Draw a free body diagram showing the forces acting on the body. That will answer your question. $\endgroup$
    – Bill N
    Jan 17 at 3:41

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(I THINK) yes, you would need to take into account that gravity is still doing work on the body if you only want to find the force caused only by the sand. I could be wrong though... as I think that the person who wrote this wants you to find the TOTAL force exerted on the body when it is in the sand.

How I would approach this question, is to first: A- assume that the sand applied a constant force to the object. Otherwise this question is impossible without knowing the exact composition of the material

Secondly, I would use conservation of energy,

mgh= 1/2mv^2

Where

m is mass of body

g is 9.81

h is 3

This is the kinetic energy of the ball once it reaches the sand.

I would then model the motion of the ball like this

$1/2mv^2 + \int_{0}^{3*10^-2} (F_{s}+F_{g}) dx = 0$

Aka, as the ball moves with some kinetic energy, the sand + gravity is going to be doing negative/positive work on the body, causing it to reach a ke of 0 once it moves 3cm

For a constant force the second term reduces to $F_{s} * 3*10^-2 + F_{g} * 3*10^-2$

Simply then rearrange to find the total force $F_{s}$

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  • $\begingroup$ The total (net) force is $F_s+F_g$. And is actually $F_s - F_g$ if you take positive up. But no matter how you choose the positive the terms have opposite signs. You want to find the $F_s$ and not the total force, which you already found from the work-energy theorem (or from N's second law,as in Steeven's answer). $\endgroup$
    – nasu
    Jan 16 at 15:29
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Now My question is if the answer is correct cause when the body will reach on the surface the gravity will still work. So, do I need to add 9.8 m/s^2 in the retardation.

This is a straight forward application of the work energy principle which states that the net work done on an object equals its change in kinetic energy. The work done by the sand in stopping the body equals its average force $F_{ave}$ times the displacement $d$ of the sand. Then, applying the principle where $m$ is the mass of the body and $v$ its velocity at impact,

$$F_{ave}d=-\frac{1}{2}mv^2$$

But since the kinetic energy at impact equals the loss in gravitational potential energy upon impact, we have equivalently

$$F_{ave}d=-mgh$$

Where $h$ is the height of the drop, ignoring the 3 cm displacement of the sand. Plugging in the data gives

$F_{ave}$= -9.8 N

as you have.

If you include the loss of potential energy associated with the penetration into the sand, which you refer to as the work gravity does after impacting the surface, then substitute $h+d$ for $h$. Or $h$=3.03 m instead of $h$=3 m. Then you get

$F_{ave}$ = -9.898 N.

Hope this helps.

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  • $\begingroup$ This means that the g should be added in the retardation because 9.898 ×10^2 = 989.8 which comes out if we add 9.8 on 980. Am i right? $\endgroup$
    – MWD
    Jan 17 at 14:00
  • $\begingroup$ I'm not sure what you mean by "adding g to the retardation". What is "retardation"? $\endgroup$
    – Bob D
    Jan 17 at 17:21
  • $\begingroup$ 980m/s^2. Retardation is (as much as i know) negative of acceleration. So either, subtracting from acceleration or adding to retardation. $\endgroup$
    – MWD
    Jan 18 at 13:22
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    $\begingroup$ OK I see what you are doing now. It is the equivalent of the work energy principle in my answer except you calculate deceleration and then force from F=ma. So, yes, you are right. But it seems like more work to calculate the velocity at impact and ensuing deceleration, than to simply use mgh. $\endgroup$
    – Bob D
    Jan 18 at 13:48
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The $a$ already includes any effect of gravity as far as I can see. So no need to add $g$.

Both in the air and in the sand, what you have to input in your motion equation is the actual acceleration.

  • In the air, that acceleration can be measured to be $g$.
  • In the sand, the acceleration is calculated from the effects from the surroundings. Meaning, $a$ is calculated under influence of both the frictional forces by the sand and gravity.

If you are not fully convinced, then try to repeat the calculation but do it over the last meter of the fall before impact with the sand. Calculate the speed that is reached after the first 2 metres of falling, and then find the acceleration $a$ in the same way as above over the final metre of the fall. If that $a$ turns out to be equal to $a=g$, then gravity was already included. If it turns out to be $a=0$, then gravity was not included and $g$ must be added to the result.

Edit:

As a comment mentions, while my answer here explains that there's no reason to alter the value of $a$, be aware that what the given question text asks for is specifically the force exerted by the sand. What you have found is the net force. So, in order to finalise the answer, you must subtract the force of gravity from the net force in order to get just the force by the sand.

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    $\begingroup$ But the question is the force exerted by the sand only? So why would the net acceleration when it is in the sand be caused the force exerted by only the sand. As you have said yourself that the acceleration is caused by the net forces of the sand and gravity? $\endgroup$ Jan 16 at 14:01
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    $\begingroup$ @jensenpaull I do see your point. But I don't think it fully makes sense to think in "partial accelerations" as you are indicating here. There is only one acceleration. Several forces cause one acceleration. Each force does not cause an acceleration which are then added up. There is no physical law that states that. We only Newton's 2nd law which tells us that the net force causes an acceleration. I would therefore never try to solve for how "much of the acceleration" one force causes. That would be very, very unusual and frankly not strictly physically correct. $\endgroup$
    – Steeven
    Jan 16 at 14:27
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    $\begingroup$ @Steeven The force of the sand needs to both decelerate the mass, and compensate for gravity. So you need to add gravity to the calculated result (or add gravitational acceleration to the computed deceleration). Think of it this way: if the force of the sand is equal to gravity, then the mass will not decelerate. $\endgroup$
    – fishinear
    Jan 16 at 15:00
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    $\begingroup$ But the question is what force has the sand exerted on the body, not the net acceleration $\endgroup$ Jan 16 at 15:00
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    $\begingroup$ @fishinear I agree with all you've said. $\endgroup$
    – Steeven
    Jan 16 at 15:01
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The force of gravity acting on the body is much smaller than the force with which the sand acts on the body, so one can neglect the gravity during the body's travel through the sand.

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