An object is moving with an initial velocity of 9m/s. It accelerates at a rate 1.5m/sec2 over a distance of 20m. What is it’s ne
1 answer:
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Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
where
giving
The rotational part requires the moment of inertia of a solid cylinder
Then the rotational kinetic energy is
Adding the two types of energy and factoring out common terms gives
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.
Kinetic energy due to linear motion is:
where
giving
The rotational part requires the moment of inertia of a solid cylinder
Then the rotational kinetic energy is
Adding the two types of energy and factoring out common terms gives
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.
Answer:
a) μ = 0.1957 , b) ΔK = 158.8 J , c) K = 0.683 J
Explanation:
We must solve this problem in parts, one for the collision and the other with the conservation of energy
Let's find the speed of the wood block after the crash
Initial moment. Before the crash
p₀ = m v₁₀ + M v₂₀
Final moment. Right after the crash
pf = m + M v_{2f}
The system is made up of the block and the bullet, so the moment is preserved
p₀ = pf
m v₁₀ = m v_{1f} + M v_{2f}
v_{2f} = m (v₁₀ - v_{1f}) / M
v_{2f} = 4.5 10-3 (400 - 190) /0.65
v_{2f} = 1.45 m / s
Now we can use the energy work theorem for the wood block
Starting point
Em₀ = K = ½ m v2f2
Final point
Emf = 0
W = ΔEm
- fr x = 0 - ½ m v₂₂2f2
The friction force is
fr = μN
With Newton's second law
N- W = 0
N = Mg
We substitute
-μ Mg x = - ½ M v2f2
μ = ½ v2f2 / gx
Let's calculate
μ = ½ 1.41 2 / 9.8 0.72
μ = 0.1957
b) let's look for the initial and final kinetic energy
K₀ = 1/2 m v₁²
K₀ = ½ 4.50 10⁻³ 400²
K₀ = 2.40 10² J
Kf = ½ 4.50 10⁻³ 190²
Kf = 8.12 10¹ J
Energy reduction is
K₀ - Kf = 2.40 10²- 8.12 10¹
ΔK = 158.8 J
c) kinetic energy
K = ½ M v²
K = ½ 0.650 1.45²
K = 0.683 J