Newton's laws

soha · 2010-05-19 21:01:47

A stone of mass 0.1 kg falls from a height of 6 m on a sandy bed and penetrates through 8 cm of sand. Calculate the average force offered by the sand.

ZHero · 2010-05-19 21:55:58

Hi soha...

Part-1: To calculate the Velocity of the stone just before it starts penetrating the sandy bed (v).

Velocity of the stone when it just starts penetrating the sandy bed can be calculated using v[sup]2[/sup] - u[sup]2[/sup] = 2a[sub]g[/sub]s

where v = final velocity of the stone (when it just starts penetrating the sandy bed)

u = initial velocity of the stone = 0 m/s (coz nothing's mentioned except that it just falls)

a[sub]g[/sub] = acceleration due to gravity = 9.8 m/s[sup]2[/sup]

s = distance covered = 6 m (distance covered while falling.. given)

Part-2: To calculate the Acceleration due to sandy bed (a[sub]s[/sub]).

Now use the relation v[sup]2[/sup] - u[sup]2[/sup] = 2a[sub]s[/sub]s

here v = final velocity of the stone = 0 m/s (coz it comes to rest after penetrating 8cm)

u = initial velocity of stone (when it just starts penetrating the sandy bed) = the value of v that you will get from Part-1.

a[sub]s[/sub] = acceleration due to sandy bed

s = 8 cm = 0.08 m (distance penetrated.. given)

Part-3: To calculate the Force exerted by the sandy bed (F).

Now use F = M × a[sub]s[/sub]

where F = force exerted by sandy bed (you have to find out)

M = mass of stone = 0.1 kg (given)

a[sub]s[/sub] = acceleration due to sandy bed = value of a[sub]s[/sub] that you got from Part-2.

I'm not gonna solve it fully for you

Hopefully that helps...

Let know if you still have any problems

Last edited by ZHero (2010-05-20 04:25:55)

soha · 2010-05-20 00:08:45

Is there any shortcut method.. Since such problems have to be solved in half minute.

soha · 2010-05-20 00:10:32

I have many such physics related doubts. Can I post them here at Mathsis fun???

While solving the first part by your method, do we get

is that correct?

Last edited by soha (2010-05-20 00:14:26)

ZHero · 2010-05-20 00:21:50

soha wrote:

I have many such physics related doubts. Can I post them here at Mathsis fun???
While solving the first part by your method, do we get
is that correct?

Just a small error...
its ∼ 10.8 (m/s)

plz recheck the decimal point.....

And why of course YES! You can always post your maths/physics problems here (MIF should agree with me here ).

Somebody will always be here to help!!

Last edited by ZHero (2010-05-20 00:45:56)

soha · 2010-05-20 00:30:46

got my mistake!!

can i post physics queries here?

ZHero · 2010-05-20 00:31:38

Oh yeahh.... I have just edited my previous post!

ZHero · 2010-05-20 00:54:40

SHORTCUT:

From the Law of Conservation of Energy....

mgh = (1/2)mv[sup]2[/sup] = work done by the opposing force = Force × distance covered

where m = mass (0.1 kg)

g = acceleration due to gravity (9.8 m/s[sup]2[/sup]

h = height of fall (6 m)

⇒ 0.1×9.8×6 = Force×0.08

⇒ Force = (0.1×9.8×6)÷0.08

Last edited by ZHero (2010-05-20 00:58:23)

soha · 2010-05-20 01:05:49

by solving it in the method which you explained first , i got
first v=10.8

next a =6.4
F=0.64N
, which is the wrong answer as per the textbook.

were did i go wrong?

ZHero · 2010-05-20 01:11:36

soha wrote:

by solving it in the method which you explained first , i got
first v=10.8
next a =6.4
F=0.64N
, which is the wrong answer as per the textbook.
were did i go wrong?

use 10.8[sup]2[/sup] - 0[sup]2[/sup] = 2a[sub]s[/sub]0.08

this should give you a[sub]s[/sub]=734.41 m/s[sup]2[/sup]

can you tell me what exactly did you do?

Last edited by ZHero (2010-05-20 04:27:07)

soha · 2010-05-20 01:15:43

soha · 2010-05-20 01:18:50

now i got a=729

Last edited by soha (2010-05-20 01:19:36)

ZHero · 2010-05-20 01:19:12

soha wrote:

use v[sup]2[/sup] - u[sup]2[/sup] instead of v[sup]2[/sup] + u[sup]2[/sup] and plz make it 0.08 m instead of 0.8 m!!!

Last edited by ZHero (2010-05-20 04:31:30)

soha · 2010-05-20 01:23:41

116.64/0.16

both have point after 2 places. so multiplying both by 100 . point gone..

Last edited by soha (2010-05-20 01:24:48)

ZHero · 2010-05-20 01:29:09

that's correct!!

but only to a few decimal places!

that's coz you've taken v = 10.8 ∼ 10.8444

you should get Force ∼ 73.441 ∼ 73.5 with v = 10.84

soha · 2010-05-20 01:37:20

so i must have taken v as 10.84

that is two places after decimal???

ZHero · 2010-05-20 01:40:50

Right again

by using this value of v, you'll get a[sub]s[/sub] ∼ 734.41 m/s[sup]2[/sup] instead of 729 m/s[sup]2[/sup].

soha · 2010-05-20 14:35:16

One more here...

A rocket of mass

is blasted vertically upwards with an initial acceleration of .Calculate the thrust of the rocket at lift off.

Last edited by soha (2010-05-20 14:37:34)

ZHero · 2010-05-20 21:02:15

soha wrote:

One more here...
A rocket of mass
is blasted vertically upwards with an initial acceleration of
.Calculate the thrust of the rocket at lift off.

To calculate Thrust (Force) use...

Thrust = (mass × acceleration) + (weight of rocket)

weight of rocket is nothing but a downward force and needs to be overcome for the rocket to be able to move upwards.. hence is added to the required thrust)

∴ Thrust = (4 × 10[sup]4[/sup] × 5) + (4 × 10[sup]4[/sup] × 9.8)

soha · 2010-05-23 00:00:25

By this way , we get the answer as

Last edited by soha (2010-05-23 00:01:09)

Math Is Fun Forum

#1 2010-05-19 21:01:47

Newton's laws

#2 2010-05-19 21:55:58

Re: Newton's laws

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#4 2010-05-20 00:10:32

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