A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s
1 answer:
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Answer:
x = 0.0685 m
Explanation:
In this exercise we can use the relationship between work and energy conservation
W = ΔEm
Where the work is
W = F x
The energy can be found in two points
Initial. Just when the block with its spring spring touches the other spring
Em₀ = K = ½ m v²
Final. When the system is at rest
=
= ½ k₁ x² + ½ k₂ x²
We can find strength with Newton's second law
∑ F = F - fr
Axis y
N- W = 0
N = W
The friction force has the equation
fr = μ N
fr = μ W
The job
W = (F – μ W) x
We substitute in the equation
(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)
½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0
We substitute values and solve
½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0
x² 30 + 14.4 x - 1,125 = 0
x² + 0.48 x - 0.0375 = 0
We solve the second degree equation
x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2
x = [-0.48 ± 0.617] / 2
x₁ = 0.0685 m
x₂ = -0.549 m
The first result results from compression of the spring and the second torque elongation.
The result of the problem is x = 0.0685 m
Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:
Finally, you replace the values of all parameters in the previous equation for I:
The moment of inertia of the door around the hinges is 2 kgm^2