A frictionless inclined plane is 8.0 m long and rests on a wall that is 2.0 m high. How much force is needed to push a block of
ice weighing 300.0 N up the plane
1 answer:
Given Information:
Inclined plane length = 8 m
Inclined plane height = 2 m
Weight of ice block = 300 N
Required Information:
Force required to push ice block = F = ?
Answer:
Force required to push ice block = 75 N
Explanation:
The force required to push this block of ice on a inclined plane is given by
F = Wsinθ
Where W is the weight of the ice block and θ is the angle as shown in the attached image.
Recall from trigonometry ratios,
sinθ = opposite/hypotenuse
Where opposite is height of the inclined plane and hypotenuse in the length of the inclined plane.
sinθ = 2/8
θ = sin⁻¹(2/8)
θ = 14.48°
F = 300*sin(14.48)
F = 75 N
Therefore, a force of 75 N is required to push this ice block on the given inclined plane.
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Answer:
The speed at the end of the track = 27 m/s
The acceleration = 1.2 m/s²
Please find the Δx vs Δt, v vs Δt, a vs Δt
Explanation:
We have;
x = u·t + 1/2·a·t²
Where;
x = The distance = 300 m
u = The initial velocity = 0 m/s (Ball at rest)
t = The time taken = 22.4 s
Therefore;
300 = 0 + 1/2×a×22.4²
a = 2×300/22.4² = 1.19579 ≈ 1.2 m/s²
v = u + a×t
∴ v = 0 + 1.2 × 22.4 = 26.88 ≈ 27 m/s
Part of the table of values is as follows;
t, x, v
0, 0, 0
0.4, 0.095663, 0.478316
0.8, 0.382653, 0.956632
1.2, 0.860969, 1.434948
1.6, 1.530611, 1.913264
2, 2.39158, 2.39158
2.4, 3.443875, 2.869896
2.8, 4.687497, 3.348212
3.2, 6.122445, 3.826528
3.6, 7.748719, 4.304844
