The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
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Answer:
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Explanation:
Answer: Instantaneous speed.
Explanation:
Answer:
So coefficient of kinetic friction will be equal to 0.4081
Explanation:
We have given mass of the block m = 0.5 kg
The spring is compressed by length x = 0.2 m
Spring constant of the sprig k = 100 N/m
Blocks moves a horizontal distance of s = 1 m
Work done in stretching the spring is equal to 
This energy will be equal to kinetic energy of the block
And this kinetic energy must be equal to work done by the frictional force
So 
So coefficient of kinetic friction will be equal to 0.4081
