Question-216110

Tinku Tara January 27, 2025 Others 0 Comments

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Question Number 216110 by Tawa11 last updated on 27/Jan/25

Commented by mr W last updated on 28/Jan/25

this workings is ok, if we don′t think about how the particle moves from P to Q.

t h i s w o r k i n g s i s o k, i f w e {d o n}^{'} t t h i n k

a b o u t h o w t h e p a r t i c l e m o v e s f r o m

P t o Q .

Commented by mr W last updated on 28/Jan/25

Commented by Tawa11 last updated on 27/Jan/25

Hahahahaha. Please solve with the 240 sir.

Hahahahaha .

Please solve with the 240 sir .

Commented by Tawa11 last updated on 27/Jan/25

Commented by Tawa11 last updated on 27/Jan/25

Commented by Tawa11 last updated on 27/Jan/25

See the workings I saw sir. I want your workings sir.

See the workings I saw sir .

I want your workings sir .

Commented by Tawa11 last updated on 27/Jan/25

Probably, he meant 240GPa. Even this is out of range. You can just take it to be 240 sir. I need your approach sir.

Probably, he meant 240 GPa .

Even this is out of range .

You can just take it to be 240 sir .

I need your approach sir .

Commented by Tawa11 last updated on 28/Jan/25

Sir, please use your approach for both cases. without considering how it moves and also consider how it moves. It will help me to solve some other exercises by myself. thanks sir.

Sir, please use your approach for both cases .

without considering how it moves and also

consider how it moves .

It will help me to solve some other exercises by myself .

thanks sir .

Commented by Tawa11 last updated on 28/Jan/25

I usually understand your approach more sir. Thanks for being there always.

I usually understand your approach

more sir . Thanks for being there always .

Commented by mr W last updated on 28/Jan/25

we must ignore how the particle moves from P to Q, otherwise we can not solve. that means this is only a question in theoretical physics world, not in real world. as i have said the workings above is ok. but it only gives the acceleration of the particle as it is leaving the point C. but the acceleration of the particle as it is approaching the point C is different. why? as the particle is approching the point C, the direction of thefriction force is downwards, while as it is leaving the point, the direction of friction is upwards.

w e m u s t i g n o r e h o w t h e p a r t i c l e

m o v e s f r o m P t o Q, o t h e r w i s e w e

c a n n o t s o l v e . t h a t m e a n s t h i s i s

o n l y a q u e s t i o n i n t h e o r e t i c a l p h y s i c s

w o r l d, n o t i n r e a l w o r l d .

a s i h a v e s a i d t h e w o r k i n g s a b o v e i s

o k . b u t i t o n l y g i v e s t h e a c c e l e r a t i o n

o f t h e p a r t i c l e a s i t i s l e a v i n g

t h e p o i n t C . b u t t h e a c c e l e r a t i o n o f

t h e p a r t i c l e a s i t i s a p p r o a c h i n g t h e

p o i n t C i s d i f f e r e n t . w h y ? a s t h e

p a r t i c l e i s a p p r o c h i n g t h e p o i n t C,

t h e d i r e c t i o n o f t h e f r i c t i o n f o r c e i s

d o w n w a r d s, w h i l e a s i t i s l e a v i n g

t h e p o i n t, t h e d i r e c t i o n o f f r i c t i o n

i s u p w a r d s .

Commented by Tawa11 last updated on 28/Jan/25

Sir, the, k = ((240N)/(1.5m)) = 160N/m? you wrote 160 m/s sir.

Sir, the,

k = \frac{240 N}{1 . 5 m} = 160 N / m ?

you wrote 160 m / s sir .

Commented by mr W last updated on 28/Jan/25

k=160 N/m =spring constant

k = 160 N / m = s p r i n g c o n s t a n t

Answered by mr W last updated on 28/Jan/25

k=((240)/(1.5))=160 N/m Δl_1 =4−1.5=2.5 m say ∣BC∣=x Δl_2 =x−1.5 energy conservation: ((kΔl_1 ^2 )/2)−μmg cos θ (x+4)=((kΔl_2 ^2 )/2)+mg(x+4) sin θ ((k(Δl_1 ^2 −Δl_2 ^2 ))/2)=mg(x+4)(sin θ+μ cos θ) ((160(2.5^2 −(x−1.5)^2 ))/2)=5×9.81(x+4)(0.8+0.5×0.6) 16x^2 −37.209x−20.836=0 ⇒x=((37.209+(√(37.209^2 +4×16×20.836)))/(2×16)) ≈2.792 m Δl_2 =2.792−1.5=1.292 m acceleration at point C (before): a_(c1) =((kΔl_2 +mg sin θ+μmg cos θ)/m) (←) =((kΔl_2 )/m)+g(sin θ+μ cos θ) =((160×1.292)/5)+9.81(0.8+0.5×0.6) =52.13 m/s^2 ✓ acceleration at point C (after): a_(c2) =((kΔl_2 +mg sin θ−μmg cos θ)/m) (←) =((kΔl_2 )/m)+g(sin θ−μ cos θ) =((160×1.292)/5)+9.81(0.8−0.5×0.6) =46.25 m/s^2 ✓

k = \frac{240}{1.5} = 160 N / m

Δ l_{1} = 4 - 1.5 = 2.5 m

s a y ∣ B C ∣= x

Δ l_{2} = x - 1.5

e n e r g y c o n s e r v a t i o n :

\frac{k Δ l_{1}^{2}}{2} - μ m g \cos θ (x + 4) = \frac{k Δ l_{2}^{2}}{2} + m g (x + 4) \sin θ

\frac{k (Δ l_{1}^{2} - Δ l_{2}^{2})}{2} = m g (x + 4) (\sin θ + μ \cos θ)

\frac{160 (2 . 5^{2} - {(x - 1.5)}^{2})}{2} = 5 \times 9.81 (x + 4) (0.8 + 0.5 \times 0.6)

16 x^{2} - 37.209 x - 20.836 = 0

\Rightarrow x = \frac{37.209 + \sqrt{37 . 209^{2} + 4 \times 16 \times 20.836}}{2 \times 16}

\approx 2.792 m

Δ l_{2} = 2.792 - 1.5 = 1.292 m

a c c e l e r a t i o n a t p o i n t C (b e f o r e) :

a_{c 1} = \frac{k Δ l_{2} + m g \sin θ + μ m g \cos θ}{m} (\leftarrow)

= \frac{k Δ l_{2}}{m} + g (\sin θ + μ \cos θ)

= \frac{160 \times 1.292}{5} + 9.81 (0.8 + 0.5 \times 0.6)

= 52.13 m / s^{2} ✓

a c c e l e r a t i o n a t p o i n t C (a f t e r) :

a_{c 2} = \frac{k Δ l_{2} + m g \sin θ - μ m g \cos θ}{m} (\leftarrow)

= \frac{k Δ l_{2}}{m} + g (\sin θ - μ \cos θ)

= \frac{160 \times 1.292}{5} + 9.81 (0.8 - 0.5 \times 0.6)

= 46.25 m / s^{2} ✓

Commented by mr W last updated on 28/Jan/25

Commented by mr W last updated on 28/Jan/25

Commented by Tawa11 last updated on 28/Jan/25

Wow, I will study your approach sir. And moreover, his answer is not the same as yours sir. He got 46.3 ms^(−2)

Wow, I will study your approach sir .

And moreover, his answer is not the

same as yours sir .

He got 46.3 {ms}^{- 2}

Commented by Tawa11 last updated on 28/Jan/25

Ohh, probably because you took g = 10m/s^2 . I will use 9.81 m/s^2 while studying your solution.

Ohh, probably because you took g = 10 m / s^{2} .

I will use 9.81 m / s^{2} while studying your solution .

Commented by mr W last updated on 28/Jan/25

he and I don′t have different approach, but exactly the same approach!

h e a n d I {d o n}^{'} t h a v e d i f f e r e n t

a p p r o a c h, b u t e x a c t l y t h e s a m e

a p p r o a c h!

Commented by Tawa11 last updated on 28/Jan/25

But I understand your presentation more sir.

But I understand your presentation

more sir .

Commented by mr W last updated on 28/Jan/25

now you should be able to solve following additional question: at point C the particle comes to instantaneous rest for the first time. say at point D it comes to instantaneous rest again. find the acceleration of the particle at point D.

n o w y o u s h o u l d b e a b l e t o s o l v e

f o l l o w i n g a d d i t i o n a l q u e s t i o n :

a t p o i n t C t h e p a r t i c l e c o m e s t o

i n s t a n t a n e o u s r e s t f o r t h e f i r s t

t i m e . s a y a t p o i n t D i t c o m e s t o

i n s t a n t a n e o u s r e s t a g a i n . f i n d

t h e a c c e l e r a t i o n o f t h e p a r t i c l e a t

p o i n t D .

Commented by Tawa11 last updated on 28/Jan/25

I will work on it sir.

I will work on it sir .

Commented by mr W last updated on 29/Jan/25

what′s your result?

{w h a t}^{'} s y o u r r e s u l t ?

Commented by Tawa11 last updated on 29/Jan/25

Please give me some time sir, I am having exam, so, I have many things to study. Thanks for helping me sir.

Please give me some time sir,

I am having exam, so, I have many

things to study . Thanks for helping

me sir .

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