Chapter 7: Q. 38 (page 179)

The coefficient of kinetic friction between the $2.0 k g$ block in FIGURE $P 7.38$ and the table is $0.30 .$ What is the acceleration of the $2.0 k g$ block?

Short Answer

Expert verified

Acceleration of the block is $6.91 m / s^{2}$ .

Step by step solution

Given information

Coefficient of friction between $2 k g$ block and table, $μ = 0.3$

Given the figure of three blocks

Explanation

Consider the net force applied by $1 k g$ block

localid="1649647806258" $F_{1} = (1 k g) \times (9.81 m / s^{2}) = 9.81 N (righttoleft)$

The friction force applied to the block $2$

localid="1649647844745" $F_{2} = (2 k g) \times 0.3 \times (9.81 m / s^{2}) = 5.8 N (righttoleft)$

The force applied by the block $3$

localid="1649647866757" $F_{3} = (3 k g) \times (9.81 m / s^{2}) = 29.43 N (lefttoright)$

Net force o block $2$

localid="1649647898415" $F = (29.43 N) - (5.8 N) - (9.81 N) = 13.82 N$

The acceleration of the block is calculated as

localid="1649647916525" $F = m a 13.82 N = 2 k g \times a a = 6.91 m / s^{2}$

Acceleration of the block is $6.91 m / s^{2}$ .

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The coefficient of kinetic friction between the $2.0 k g$ block in FIGURE $P 7.38$ and the table is $0.30 .$ What is the acceleration of the $2.0 k g$ block?

Short Answer

Step by step solution

Given information

Explanation

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The coefficient of kinetic friction between the 2.0kgblock in FIGURE P7.38 and the table is 0.30. What is the acceleration of the 2.0kg block?

Short Answer

Step by step solution

Given information

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Nuclear Physics

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The coefficient of kinetic friction between the $2.0 k g$ block in FIGURE $P 7.38$ and the table is $0.30 .$ What is the acceleration of the $2.0 k g$ block?