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A box with mass m is dragged across a level floor with coefficient of kinetic friction μk by a rope that is pulled upward at an angle θ above the horizontal with a force of magnitude F. (a) In terms of m, μk, θ, and g , obtain an expression for the magnitude of the force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90-kg patient across a floor at constant speed by pulling on him at an angle of 25∘ above the horizontal. By dragging weights wrapped in an old pair of pants down the hall with a spring balance, you find that μk=0.35. Use the result of part (a) to answer the instructor’s question.

Respuesta :

Answer:

a) F =  μk  mg Cosθ

b) F = 279.78 N

Explanation:

a) F = μk R

Based on the description in the question, the horizontal reaction is:

R = mg Cosθ

The force required to move the box with constant speed in terms of m, μk, θ, and g is :

F =  μk  mg Cosθ

b) If m = 90 kg

g = 9.8 m/s²

μk=0.35

θ = 25⁰

Force required to slide the 90-kg patient across a floor at constant speed by pulling on him at an angle of 25∘ above the horizontal will be:

F =  μk  mg Cosθ

F = 0.35 * 90 * 9.8 * cos25

F = 279.78 N

a) The expression for the magnitude of the force is  [tex]F = \mu\ k \ mg Cos\theta[/tex]

b)The force is F = 279.78 N

  • The calculation is as follows:

a) [tex]F = \mu k R[/tex]

Depend on the description in the question, the horizontal reaction is:

[tex]R = mg Cos\theta[/tex]

The force needed to move the box with constant speed in terms of m, [tex]\mu k, \theta[/tex], and g is :

[tex]F = \mu\ k \ mg Cos\theta[/tex]

b)

Now If m = 90 kg

g = 9.8 m/s²

[tex]\mu k=0.35\\\\\theta = 25^{\circ}[/tex]

Now

[tex]F = \mu\ k \ mg Cos\theta[/tex]

[tex]F = 0.35 \times 90 \times 9.8 \times cos25[/tex]

F = 279.78 N

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