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vlabodo [156]
3 years ago
5

A girl pulls her younger brother on a sled along a flat sidewalk (the total mass of the sled is 30kg). A frictional force of 50N

opposes the pull. When moving, the girl pulls on the sled with a force of 150N at an angle of 30 degrees.
How much time will it take for the sled to travel 50?

What is the normal force?
Physics
1 answer:
harkovskaia [24]3 years ago
4 0

Answer:

6.13 s

219 N

Explanation:

Newton's law in the x direction:

∑F = ma

150 cos 30° N − 50 N = (30 kg) a

a = 2.66 m/s²

Δx = v₀ t + ½ at²

(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²

t = 6.13 s

Newton's law in the y direction:

∑F = ma

Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0

Fn = 219 N

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Answer:

a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

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Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension

the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension

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B

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Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂ + x₃

x = F/k = F/k₁ + F/k₂ + F/k₃

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\text { Distance }(\mathrm{d})=\text { speed }(\mathrm{s}) \times \text { time }(\mathrm{t})

\text { speed }=\frac{\text {distance}}{\text {time}}

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\text { Speed }=10 \mathrm{km} \times 12 \mathrm{hour}

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