Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK = 
-K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W -
= -K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3
]
v = √(31.63 2/4.3)
v = 3.84 m/s
The best position for the person would be outside, under a clear sky, standing up. He should do it sometime between sunset and sunrise, from a day before until a day after the moment of Full Moon.
Answer:
$ 0.48
Explanation:
We can calculate this quantity easily using successive products and taking into account the units.
![\frac{0.08}{kw*h}*2[kw]*3[hr]\\ \\=0.48](https://tex.z-dn.net/?f=%5Cfrac%7B0.08%7D%7Bkw%2Ah%7D%2A2%5Bkw%5D%2A3%5Bhr%5D%5C%5C%20%5C%5C%3D0.48)
The amount is $ 0.48
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.
The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds) = 4,000 foot-pounds of work.
If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds) = 645.2 foot-pounds per second.
The rate of doing work is called "power".
(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)
So back to our problem.
John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.
Well, 550 foot-pounds per second is called 1 "horsepower".
So as John runs up the steps to the balcony, he's doing the work
at the rate of
(645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)
= 1.173 Horsepower. GO JOHN !
(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________
Oh my gosh ! Look at #26 ! There are the metric units I was talking about.
Do you need #26 ?
I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.
a). 5
b). 750 Joules
c). 800 Joules
d). 93.75%
You're welcome.
And #27 is 0.667 m/s .
