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2 years ago

A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.

Physics

1 answer:

sasho [114]2 years ago

7 0

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

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2 years ago

A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. T

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Answer:

Option d)

Solution:

As per the question:

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Now,

The component of force acting horizontally is F'cos $\theta$

Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.

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<h2>Solution :</h2>

Here ,

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• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

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