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3 years ago

6

A 2.0-kg silverware drawer does not slide readily. The owner gradually pulls with more and more force, and when the applied forc

e reaches 9.0 N, the drawer suddenly opens, throwing all the utensils to the floor. What is the coefficient of static friction between the drawer and the cabinet?

1 answer:

topjm [15]3 years ago

7 0

Answer: 0.45

Explanation:

First note that the body that causes the body to move is its moving force (Fm) which is 9.0N

Since the mass of the body is 2.0kg, the weight will be;

W= mg = 2×10

W= 20N

For static body, the frictional force (Ff) acting on the body is equal to the moving force (Fm) since both forces acts along the horizontal on the body.

Ff = Fm = 9.0N

The normal reaction (R) on the body will also be equal to its weight(W) since weight acts downwards and the reaction acts in the opposite direction (upwards).

R = W = 20N

Ff = nR taking 'n' as coefficient of static friction between the drawer and the cabinet.

9.0 = 20n

n = 9/20

n = 0.45

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<em>where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$ : are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$ : are the proton and helium nucleus velocities, respectively, after the collision </em>

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$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

<em>where $K_{1i}$ and $K_{2i}$ : are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and $K_{2f}$ : are the kinetic energy for the proton and helium, respectively, after the colission </em>

<u>From (3):</u>

$\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$ (4)

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Solving equation (2) for $v_{1f}$ , we have:

$v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f}$ (5)

<u>From getting (5) into (4) we can obtain the $v_{2f}$ :</u>

$v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0$

$v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0$

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

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Now, by introducing (6) into (5) we get the proton velocity after the collision:

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$v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}$

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

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