SATHEE: Work Energy and Power - Result Question 29

Work Energy and Power - Result Question 29

32. Consider a car moving along a straight horizantal road with a speed of $72 k m / h$ . If the coefficient of static friction between road and tyres is 0.5 , the shortest distance in which the car can be stopped is

[1994]

(a) $30 m$

(b) $40 m$

(d) $20 m$

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Answer:

Correct Answer: 32. (b)

Solution:

(b) Force due to friction $=$ kinetic energy

$μ m g s = \frac{1}{2} m v^{2}$

$[.$ Here, $v = 72 k m / h = \frac{72000}{60 \times 60} = 20 m / s$ ] or , $s = \frac{v^{2}}{2 μ g} = \frac{20 \times 20}{2 \times 0.5 \times 10} = 40 m$

If a body of mass $m$ moves with velocity $u$ on a rough surface and stops after travelling a distance $S$ due to friction. Then, frictional force, $F = m a =$ $μ R \Rightarrow m a = μ m g \Rightarrow a = μ g$

Using $v^{2} = u^{2} - 2 a s$

$\Rightarrow 0 = u^{2} - 2 μ g S$

$\Rightarrow S = \frac{u^{2}}{2 μ g}$

Work Energy and Power - Result Question 29

32. Consider a car moving along a straight horizantal road with a speed of $72 k m / h$ . If the coefficient of static friction between road and tyres is 0.5 , the shortest distance in which the car can be stopped is

Answer:

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Work Energy and Power - Result Question 29

32. Consider a car moving along a straight horizantal road with a speed of 72km/h. If the coefficient of static friction between road and tyres is 0.5 , the shortest distance in which the car can be stopped is

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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32. Consider a car moving along a straight horizantal road with a speed of $72 k m / h$ . If the coefficient of static friction between road and tyres is 0.5 , the shortest distance in which the car can be stopped is