What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? Vie
2 answers:
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Answer:
μ = 0.0315
Explanation:
Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)
N = m*g (1)
The friction force is equal to the product of the normal force by the coefficient of friction.
F = μ*N (2)
This way replacing 1 in 2, we have:
F = μ*m*g (2)
Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.
where:
vf = final velocity = 0
vi = initial velocity = 85 [km/h] = 23.61 [m/s]
d = displacement = 900 [m]
F = friction force [N]
The final velocity is zero since when the vehicle has traveled 900 meters its velocity is zero.
Now replacing:
(1/2)*m*(23.61)^2 = μ*m*g*d
0.5*(23.61)^2 = μ*9,81*900
μ = 0.0315