On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1
1 answer:
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Answer:
(a) the coefficient of friction is 0.451
This was derived using the energy conservation principle (the total energy in a closed system remains constant).
(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.
Explanation:
For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.
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1) For x = 6.6 cm,
2) For x = 6.6 cm,
3) For x = 1.45 cm,
4) For x = 1.45 cm,
5) Surface charge density at b = 4 cm:
6) At x = 3.34 cm, the x-component of the electric field equals zero
7) Surface charge density at a = 2.9 cm:
8) None of these regions
Explanation:
1)
The electric field from an infinite charge sheet is perpendicular to it:
where
is the surface charge density
represents vacuum permittivity
Outside the slab, the electric field behaves like that of an infinite sheet.
Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:
where
The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).
Thus,
and the negative sign indicates a rightward direction.
2)
Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.
<pThus, the y-component totals zero.This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, , will be equal and opposite to the corresponding component from the opposite side,
. Thus, the combined y-direction field is always zero.
3)
This scenario resembles part 1), but the point here is
x = 1.45 cm
which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to
Replacing with expressions from part 1), we get
where the negative illustrates a leftward direction.
4)
This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.
5)
Notably, the slab behaves as a conductor, signifying charge mobility within it.
The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).
The surface charge density per unit area of the slab is
This average denotes the surface charge density on both slab sides at points a and b:
(1)
Additionally, the infinite sheet at x = 0 negatively charged , induces an opposite net charge on the slab's left surface, thus
(2)
Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:
6)
We aim to compute the x-component of the electric field at
x = 3.34 cm
This point lies inside the slab, bounded at
a = 2.9 cm
b = 4.0 cm
In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero
7)
From part 5), we determined the surface charge density at x = a = 2.9 cm is
8)
As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are
2.9 cm < x < 4 cm
Thus, the suitable answer is
"none of these regions"
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