Answered: The bottom crate is pulled by a force…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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The bottom crate is pulled by a force of 20 N to the right. The μk between the bottom crate and the surface is 0.32. The μk between the bottom crate and the top crate is also 0.32.
(Figure 1) (below)

Find the acceleration of the 2.0 kg crate.

Express your answer to two significant figures and include the appropriate units.

Expert Solution

Step 1

Given data:

The force acting on the bottom crate is F=20 N.
The coefficient of friction between bottom crate and the surface is μ_k=0.32.
The coefficient of friction between bottom crate and top crate is μ_k=0.32.
The mass of the upper crate is m₁=1 kg.
The mass of the bottom crate is m₂=2 kg.

The representation of various forces acting on upper crate is shown as,

Step 2

The expression for the tension in the string due to upper crate can be calculated as,

$T = μ_{k} N_{1} + m_{1} a T = (μ_{k} m_{1} g) + m_{1} a$

Here, N₁ is the normal force of upper crate, and g is the gravitational acceleration having standard value of 9.81 m/s².

Substitute the known values in the above expression.

$T = (0.32 \times 1 kg \times 9.81 m / s^{2}) + (1 kg \times a) T = 3.14 kg \cdot m / s^{2} + (1 kg \times a) . . . . . . . . . (1)$

The representation of various forces acting on bottom crate is shown as,

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