1. A block of mass m = 10.0 kg is released with a speed v from a frictionless incline at height 7.00 m. The block reaches the horizontal ground and then slides up another frictionless incline as shown in Fig. 1.1. If the
horizontal surface is also frictionless and the maximum height that the block can slide up to is 26.0 m, (a) what
is the speed v of the block equal to when it is released and (b) what is the speed of the block when it reaches
the horizontal ground? If a portion of length 1 2.00 m on the horizontal surface is frictional with coefficient
of kinetic friction uk = 0.500 (Fig. 1.2) and the block is released at the same height 7.00 m with the same
speed v determined in (a), (c) what is the maximum height that the block can reach, (d) what is the speed of the
block at half of the maximum height, and (e) how many times will the block cross the frictional region before
it stops completely?
1 = 2.00 m (frictional region)

Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).

At point A, the block has total energy

E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²

E (A) = 686 J + 1/2 (10.0 kg) v₀²

At point B, the block's potential energy is converted into kinetic energy, so that its total energy is

E (B) = 1/2 (10.0 kg) v₁²

The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,

E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J

Throughout this whole process, energy is conserved, so

E (A) = E (B) = E (C) = E (D)

(a) Solve for v₀ :

686 J + 1/2 (10.0 kg) v₀² = 2548 J

==>   v₀ ≈ 19.3 m/s

(b) Solve for v₁ :

1/2 (10.0 kg) v₁² = 2548 J

==>   v₁ ≈ 22.6 m/s

Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:

• net horizontal force:

∑ F = -f = ma

• net vertical force:

∑ F = n - mg = 0

where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :

n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N

f = µn = 0.500 (98.0 N) = 49.0 N

==>   - (49.0 N) = (10.0 kg) a

==>   a = - 4.90 m/s²

The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that

v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)

==>   v₂² = 490 m²/s²

and thus the block has total/kinetic energy

E (C) = 1/2 (10.0 kg) v₂² = 2450 J

(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so

2450 J = (10.0 kg) (9.80 m/s²) h

==>   h = 25.0 m

(d) At half the maximum height, the block has speed v₃ such that

2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²

==>   v₃ ≈ 15.7 m/s

The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by

v = v₁ + at = 22.6 m/s - (4.90 m/s²) t

The block comes to a rest when v = 0 :

0 = 22.6 m/s - (4.90 m/s²) t

==>   t ≈ 4.61 s

It covers a distance x after time t of

x = v₁t + 1/2 at ²

so when it comes to a complete stop, it will have moved a distance of

x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m

(e) The block crosses the rough region

(52.0 m) / (2.00 m) = 26 times

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).