A train is 240 meters long and travels 20 m/s. How long does
1 answer:
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Answer:
100 J
Explanation:
Given:
Force acting on the crate is,
Displacement of the crate is,
Angle between the direction of force and displacement is,
Now, work done by a force 'F' causing a displacement 'S' such that the angle between the force and displacement is is given as:
Now, plug in 100 N for 'F', 2 m for 'S', and 60° for . This gives,
Therefore, the work done by the force is 100 J.
Answer:
μ = 0.498
Explanation:
I manage to find the picture of this problem. In the attached picture you can also see the forces involved in this case.
We know according to newton's law that the static friction force is:
Fs = μ * N
However, as you can see in the picture, the Normal force is equals to the weight, in this case the weight of the painter and also the ladder, therefore:
N = Fy
and the force of the friction is:
Fs = Fx
Therefore the coefficient is:
μ = Fx/Fy (1)
Now, let's write the equations in x and y, to solve this.
For the "x" axis:
Fx - Fw = 0 -----> Fx = Fw (2)
Fw is force of the wall, while Fx is the force friction in the x axis (base of the ladder).
for the "y" axis:
Fy - W1 - W2 = 0
W1 = mg (ladder)
W2 = Mg (painter)
replacing we have:
Fy = W1 + W2
Fy = mg + Mg ----> Fy = g(m + M) (3)
To get the force that the wall is exerting we need to calculate the torque around the foot of the ladder so:
τ = rF sinθ
However, the angle in the wall and the ladder is 90° so:
τ = rF sin90°
τ = rF (4)
replacing (4) with the forces we have:
rFw = rW1 + rW2
4Fw = 1.5mg + 2.1Mg
Fw = 1.5mg + 2.1Mg/4 (5)
Finally, with this expression, we can replace it in (1) to get the coefficient of friction:
μ = Fx/Fy
μ = 1.5mg + 2.1Mg / 4g(m + M) gravity cancels out so:
μ = 1.5m + 2.1M / 4(m + M)
Replacing the data we finally have:
μ = (1.5 * 2) + (2.1 * 55) / 4 (55 + 12)
μ = 133.5 / 268
<h2>μ = 0.498</h2>