SATHEE: Laws of Motion - Result Question 47

Laws of Motion - Result Question 47

47. A block of mass $1 k g$ is placed on a truck which accelerates with acceleration $5 m / s^{2}$ . The coefficient of static friction between the block and truck is 0.6 . The frictional force acting on the block is

(a) $5 N$

(b) $6 N$

(d) $4.6 N$

[2001]

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Answer:

Correct Answer: 47. (a)

Solution:

(a) Maximum friction force $= μ m g$

$= .6 \times 1 \times 9.8 = 5.88 N$

But here required friction force

$= m a = 1 \times 5 = 5 N$

For a body placed on a moving body and Ist body not to slide - applied force $F <$ limiting friction ( $=$ Ms mg) combined system will move together with common acceleration $a_{1} = a_{2} = \frac{F}{M + m}$ .

Laws of Motion - Result Question 47

47. A block of mass $1 k g$ is placed on a truck which accelerates with acceleration $5 m / s^{2}$ . The coefficient of static friction between the block and truck is 0.6 . The frictional force acting on the block is

Answer:

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Laws of Motion - Result Question 47

47. A block of mass 1kg is placed on a truck which accelerates with acceleration 5m/s2. The coefficient of static friction between the block and truck is 0.6 . The frictional force acting on the block 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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47. A block of mass $1 k g$ is placed on a truck which accelerates with acceleration $5 m / s^{2}$ . The coefficient of static friction between the block and truck is 0.6 . The frictional force acting on the block is