Question

Question for Physics Lovers !

=> A 1kg block situated on a rough incline is connected to a spring of negligible mass having spring constant 100N/m as shown in the attachment.
=> The block is released from rest with the spring in the unstretched position. The block moves 10cm down the incline before coming to rest. The coefficient of friction between the block and the incline is
(Take g = 10m/s² and assume that the pulley is frictionless)

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Answer 1

Given :

▪ Mass of block = 1kg

▪ Angle of inclination = 45°

▪ Spring constant = 100N/m

To Find :

▪ Co-efficient of friction b/w the block and the incline.

SoluTion :

✴ From figure,

$\dashrightarrow\tt\:N=mg\cos\theta\\ \\ \dashrightarrow\tt\:\blue{f=\mu N=\mu mg\cos\theta}$

→ where $\mu$ is the coefficient of friction between the block and the incline.

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→ Net force on the block down the incline,

$\Rightarrow\tt\:F_{net}=mg\sin\theta -f\\ \\ \Rightarrow\tt\:F_{net}=mg\sin\theta-\mu mg\cos\theta\\ \\ \Rightarrow\tt\:\pink{F_{net}=mg(\sin\theta -\mu\cos\theta)}$

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→ Distance moved, x = 0.1m

In equilibrium,

$\twoheadrightarrow\sf\:\green{Work\:done=Potential\:energy\:of\:stretched\:spring}\\ \\ \twoheadrightarrow\tt\:mg(\sin\theta-\mu\cos\theta)x=\dfrac{1}{2}kx^2\\ \\ \twoheadrightarrow\tt\:2mg(\sin\theta-\mu\cos\theta)=kx\\ \\ \twoheadrightarrow\tt\:2\times 1\times 10\times (\sin 45\degree-\mu\cos 45\degree)=100\times 0.1\\ \\ \twoheadrightarrow\tt\:\sin45\degree-\mu\cos45\degree=\dfrac{1}{2}\\ \\ \twoheadrightarrow\tt\:\dfrac{1}{\sqrt{2}}-\dfrac{\mu}{\sqrt{2}}=\dfrac{1}{2}\\ \\ \twoheadrightarrow\tt\:1-\mu=\dfrac{1}{\sqrt{2}}\\ \\ \twoheadrightarrow\tt\:\mu=\dfrac{\sqrt{2}-1}{\sqrt{2}}\\ \\ \twoheadrightarrow\underline{\boxed{\bold{\tt{\red{\mu=0.3}}}}}\:\orange{\bigstar}$

Answer 2

Given ,

Mass of block , m = 1 kg

Spring constant , k = 100 N/m

Distance moved by block , x = 10 cm = 0.1 m

Angle , ∅ = 45⁰

Gravity , g = 10 m/s²

( Assume that the pulley is frictionless )  

In y - direction :-

$\bold{F_y=N-mgcos\theta}\\\\\bold{0=N-mgcos\theta}\\\\\bold{Since\;net\;force\;along\;y-axis\;is\;zero}\\\\\bold{N=mgcos\theta...(1)}$

In x - direction :-

$\bold{F_y=mgsin\theta-friction}\\\\\bold{Now\;at\;equilibrium\;condition\;work\;done\;is\;equals\;to\;spring\;force}\\\\\bold{\frac{1}{2}kx^2=(mgsin\theta-uN)x}$Since Spring force is acting along x-axis

$\bold{\frac{1}{2}kx^2=(mgsin\theta-u.mgcos\theta)x\;[From\;(i)] }\\\\\bold{kx=2mg(sin\theta-u.cos\theta)}\\\\\bold{100*(0.1)=2*1*10[sin45-u.cos45]}\\\\\bold{10=\frac{20}{\sqrt{2}}[1-u] }\\\\\bold{u=1-\frac{1}{\sqrt{2}} }\\\\\bold{\bf{\blue{u=0.3}}}$

So the coefficient of friction between the block and incline is 0.3