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miss Akunina
1 month ago
11

A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above

. If FP remains constant but the angle θ is increased, which of the following is true at some later time?
A.The force of friction between the block and surface will increase.

B.The force of friction between the block and surface will decrease.

C.The weight of the block will increase.

D.The weight of the block will decrease.

Physics
2 answers:
serg [3.5K]1 month ago
6 0

Response:

B. The friction force between the block and surface will diminish.

Clarification:

The force of friction can be expressed as

F_s = \mu N

where \mu represents the friction coefficient and N signifies the normal force.

As the student pulls the block with force F_p at an angle \theta, the normal force acting on the block subsequently becomes

N = Mg- F_psin(\theta)

and therefore, the frictional force changes to

F_s = \mu (Mg- F_psin(\theta)).

With an increase in \theta, the sin(\theta) increases, which subsequently reduces the normal force Mg- F_psin(\theta), indicating a decrease in the frictional force; hence choice B is correct.

Note: While choice D seems appealing, it is incorrect since the weight W=mg remains unaffected by the external forces acting on the block.

Sav [3.1K]1 month ago
3 0
The frictional force between the block and the surface will lessen.
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A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b
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Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b,   noting that r = b

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C. At r = a

Following from equation 1:

E(r) = 0 for r≤a.  Equation 1

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The electric potential can be determined by

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As r approaches 0,

V = kq / 0

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The solution leads to the conclusion that m1 = m2

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