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Mary pushes a crate by applying a force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they apply a force of 48 newtons, and the crate just starts moving. If the coefficient of static friction is 0.11, what is the value of the normal force?

A. [tex]$2.0 \times 10^2$[/tex] newtons
B. [tex]$2.6 \times 10^2$[/tex] newtons
C. [tex]$3.9 \times 10^2$[/tex] newtons
D. [tex]$4.0 \times 10^2$[/tex] newtons
E. [tex]$4.3 \times 10^2$[/tex] newtons

Sagot :

GinnyAnswer
Let's go through the problem step-by-step to determine the normal force.

1. First, we know that Mary and Anne together apply a total force of [tex]\(48\)[/tex] newtons to just start moving the crate. This force corresponds to the force of static friction that they need to overcome.

2. The coefficient of static friction ([tex]\(\mu\)[/tex]) is given as [tex]\(0.11\)[/tex].

3. The equation that relates the force of static friction ([tex]\(f_{friction}\)[/tex]) to the normal force ([tex]\(N\)[/tex]) and the coefficient of static friction is:
[tex]\[ f_{friction} = \mu \times N \][/tex]

4. Rearranging this equation to solve for the normal force ([tex]\(N\)[/tex]), we get:
[tex]\[ N = \frac{f_{friction}}{\mu} \][/tex]

5. Substituting the given values into this equation:
[tex]\[ N = \frac{48 \text{ newtons}}{0.11} \][/tex]

6. Calculating the above expression gives us:
[tex]\[ N \approx 436.36 \text{ newtons} \][/tex]

Reviewing the answer choices:
A. [tex]\(2.0 \times 10^2\)[/tex] newtons = 200 newtons
B. [tex]\(2.6 \times 10^2\)[/tex] newtons = 260 newtons
C. [tex]\(3.9 \times 10^2\)[/tex] newtons = 390 newtons
D. [tex]\(4.0 \times 10^2\)[/tex] newtons = 400 newtons
E. [tex]\(4.3 \times 10^2\)[/tex] newtons = 430 newtons

The closest value to [tex]\(436.36\)[/tex] newtons is [tex]\(4.3 \times 10^2\)[/tex] newtons.

Therefore, the correct answer is:
E. [tex]\(4.3 \times 10^2\)[/tex] newtons
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