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3 years ago

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Jonah is trying to move his 22-kg desk by pushing on it with a force of 130 N, but his brother is leaning on it with a downward

force of 24 N, just preventing it from moving. What is the coefficient of static friction between the desk and the floor?

1 answer:

Dafna11 [192]3 years ago

6 0

Answer:

0.54

Explanation:

Draw a free body diagram. There are 5 forces on the desk:

Weight force mg pulling down

Applied force 24 N pushing down

Normal force Fn pushing up

Applied force 130 N pushing right

Friction force Fnμ pushing left

Sum of the forces in the y direction:

∑F = ma

Fn − mg − 24 = 0

Fn = mg + 24

Fn = (22)(9.8) + 24

Fn = 240

Sum of the forces in the x direction:

∑F = ma

130 − Fnμ = 0

Fnμ = 130

μ = 130 / Fn

μ = 130 / 240

μ = 0.54

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Explanation:

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The image formed from the lens is real, inverted and enlarged

Explanation:

The magnification of a lens is given by:

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y is the size of the object

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Electric field is given by

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We use the sum of the masses because the objects are stuck together after the collision. The initial kinetic energy is:

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Substituting we get:

$\Delta K=\frac{1}{2}(m_1+m_2)v^2_f-(\frac{1}{2}m_1v^2_{01}+\frac{1}{2}m_2v^2_{02})$

The only missing variable is the final velocity. To determine the final velocity we will use the conservation of momentum.

We will use the conservation of momentum in the horizontal direction (west) and the conservation of momentum in the vertical direction (north).

In the horizontal direction we have:

$m_1v_{h1}+m_2v_{h2}=(m_1+m_2)v_{hf}$

Since the object 1 has no velocity in the horizontal direction we have that:

$\begin{gathered} m_1(0)+m_2v_{h2}=(m_1+m_2)v_{hf} \\ m_2v_{h2}=(m_1+m_2)v_{hf} \end{gathered}$

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Now we substitute the values:

$\frac{(6kg)(2\frac{m}{s})}{(4kg+6kg)}=v_{hf}$

Solving the operations we get:

$1.2\frac{m}{s}=v_{hf}$

Now we use the conservation of momentum in the vertical direction, we get:

$m_1v_{v1}+m_2v_{v2}=(m_1+m_2)v_{vf}$

Since the second object has no vertical velocity we get:

$m_1v_{v1}=(m_1+m_2)v_{vf}$

Now w solve for the final vertical velocity, we get:

$\frac{m_1v_{v1}}{\mleft(m_1+m_2\mright)}=v_{vf}$

Now we substitute the values:

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Now we solve the operations:

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$v_f=\sqrt[]{v^2_{hf}+v^2_{vf}}$

Substituting the values:

$v_f=\sqrt[]{(1.2\frac{m}{s})^2+(2\frac{m}{s})^2}$

Solving the operations:

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Now we substitute this in the formula for the kinetic energy and we get:

$\Delta K=\frac{1}{2}(4kg+6kg)(2.53\frac{m}{s})^2-(\frac{1}{2}(4kg)(5\frac{m}{s})^2+\frac{1}{2}(6kg)(2\frac{m}{s})^2)$

Solving the operations:

$\Delta K=32J-62J=-30J$

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