Answer:
Explanation:
a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²
500 = 5 x 10⁻³ k ,
k = (500/5) x 10³ = 10⁵ N/m
b )
k = 4.5 x 10¹ = 45 N/m
Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J
This energy gets dissipated by friction .
work done by friction = μ mg d
d is the distance traveled under friction
so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2
μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .
Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
![(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]](https://tex.z-dn.net/?f=%281000%2A10%29%2B%282000%2A0%29%3D%281000%2B2000%29%2Av_%7B3%7D%5C%5Cv_%7B3%7D%3D3.33%5Bm%2Fs%5D)
The conversion from gallons to liters is 1 = 3.785.
Keeping this in mind...
42 x 3.785 = 158.97 liters.
If rounding, there are about 159 liters of oil in a barrel.
The total momentum of the system has to be conserved to satisfy the principle of conservation of momentum. Before the ball hits the bottle, the momentum of the system is 0.4 x 18 = 7.2 kg m/s
The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s
So the momentum of the ball now is 7.2 - 5 = 2.2 kg m/s
Hence its velocity is 2.2/0.4 = 5.5 m/s
The answer is D because it’s going by the miles