SATHEE: Work Power and Energy 3 Question 6

Work Power and Energy 3 Question 6

6. A $0.5 k g$ block slides from the point $A$ (see fig.) on a horizontal track with an initial speed of $3 m / s$ towards a weightless horizontal spring of length $1 m$ and force constant $2 N / m$ . The part $A B$ of the track is frictionless and the part $B C$ has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances $A B$ and $B D$ are $2 m$ and $2.14 m$ respectively, find the total distance through which the block moves before it comes to rest completely. (Take $g = 10 m / s^{2}$ ).

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$(1983, 7 M)$

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Answer:

Correct Answer: 6. $4.24 m$

Solution:

From $A$ to $B$ , there will be no loss of energy. Now, let block compresses the spring by an amount $x$ and comes momentarily to rest. Then, loss of energy will be equal to the work done against friction. Therefore,

$μ_{k} m g (B D + x) = \frac{1}{2} m v^{2} - \frac{1}{2} k x^{2}$

Substituting the values

$(0.2) (0.5) (10) (2.14 + x) = \frac{1}{2} (0.5) (3)^{2} - \frac{1}{2} (2) (x)^{2}$

Solving this equation, we get $x = 0.1 m$

Now, spring exerts a force $k x = 0.2 N$ on the block. But to stop the block from moving limiting static friction is $μ_{s} m g = (0.22) (0.5) (10) = 1.1 N$ . Since, $1.1 N > 0.2 N$ , block will not move further and it will permanently stop there. Therefore, total distance covered before it comes to rest permanently is

$d = A B + B D + x = 2 + 2.14 + 0.1 = 4.24 m$

Work Power and Energy 3 Question 6

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Work Power and Energy 3 Question 6

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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