Answer:
1838216 J
Explanation:
95 km/h = 26.39 m/s
40 km/h = 11.11 m/s
Initial kinetic energy
= .5 x 1600 x(26.39)²
= 557145.67 J
Final kinetic energy
= .5 x 1600 x ( 11.11)²
= 98745.68 J
Loss of kinetic energy
= 458400 J
Loss of potential energy
= mg x loss of height
= 1600 x 9.8 x 340 sin 15
= 1379816 J
Sum of Loss of potential energy and Loss of kinetic energy
= 1379816 + 458400
= 1838216 J
This is the work done by the friction . So this is heat generated.
As momentum / time = force
so; time = 100÷15
so your answer is 6.7 !!
Answer:
(a) Vf = 128 ft/s
(b) K.E = 122.8 Btu
Explanation:
(a)
In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = 32.2 ft/s²
h = height = 253 ft
Vf = Final Velocity = ?
Vi = Initial Velocity = 10 ft/s
Therefore,
(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²
16293.2 ft²/s² + 100 ft²/s² = Vf²
Vf = √(16393.2 ft²/s²)
<u>Vf = 128 ft/s</u>
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(b)
The kinetic energy of the object before it hits the surface of earth is given by:
K.E = (0.5)(m)(Vf)²
where,
m = mass of object = 375 lb
K.E = Kinetic energy of object before it strikes the surface of earth = ?
Therefore,
K.E = (0.5)(375 lb)(128 ft/s)²
K.E = 3073725 lb.ft²/s²
Now, converting this to Btu:
K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)
<u>K.E = 122.8 Btu</u>
If you remember the formula for potential energy,
then this question is a piece-o-cake.
<em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>
-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ?
Compared to gravity on Earth, it's only 16.5 percent as much on the Moon.
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.
