Three perfectly polarizing sheets are spaced 2?cm {\rm cm} apart and in parallel planes. The transmission axis of the second she
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Answer:
v= 4.0 m/s
Explanation:
- When standing on the bathroom scale within the moving elevator, there are two forces acting on Henry's mass: Normal force and gravity.
- Gravity is always downward, and normal force is perpendicular to the surface on which the mass is located (the bathroom scale), in upward direction.
- Normal force, can adopt any value needed to match the acceleration of the mass, according to Newton's 2nd Law.
- Gravity (which we call weight near the Earth's surface) can be calculated as follows:
- According to Newton's 2nd Law, it must be met the following condition:
- As the gravity is larger than normal force, this means that the acceleration is downward, so, we choose this direction as the positive.
- Solving for a, we get:
- We can find the speed after the first 3.8 s (assuming a is constant), applying the definition of acceleration as the rate of change of velocity:
- Now, if during the next 3.8 s, normal force is 930 N (same as the weight), this means that both forces are equal each other, so net force is 0.
- According to Newton's 2nd Law, if net force is 0, the object is either or at rest, or moving at a constant speed.
- As the elevator was moving, the only choice is that it is moving at a constant speed, the same that it had when the scale was read for the first time, i.e., 4 m/s downward.
Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)