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drek231 [11]
4 years ago
15

Now imagine a person dragging a 50 kg box along the ground with a rope, as

Physics
1 answer:
ANTONII [103]4 years ago
7 0

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

                                                     = 50 x 9.8

                                                     = 490 N

The coefficient of friction,

                            μ = f/ η

                               = 30 / 490

                               = 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

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Answers:

a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

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V=\omega L=\frac{2\pi}{T} L (1)

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T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s (8)

Ordering this periods from largest to smallest:

T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2} (10)

a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2} (11)

a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2} (13)

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a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

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