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solniwko [45]
3 years ago
14

Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time movi

ng with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Physics
1 answer:
Leviafan [203]3 years ago
5 0

Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

<u>W = 12.96 J</u>

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Answer:

The correct option is;

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A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied
Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that  the minimum force that must be applied on the <em>book is</em>

  • F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the  Force  is mathematically given as

F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

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