A 5000 kg truck traveling at 60 m/s stops in 5 seconds. How much friction was between the truck's tires and the ground?
1 answer:
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If you illustrate the problem, you will somewhat come up with the figure shown. The missing value is the hypotenuse of the right triangle. Using the pythagorean theorems, the value is determined to be
x = √(0.7^2 + 2.4^2)
x = 2.5 km
2.5 km is the magnitude of the distance. If you want to incorporate the displacement, the answer is reported as 2.5 km, southeast. The direction is determined from the starting point to the endpoint.
x = √(0.7^2 + 2.4^2)
x = 2.5 km
2.5 km is the magnitude of the distance. If you want to incorporate the displacement, the answer is reported as 2.5 km, southeast. The direction is determined from the starting point to the endpoint.
Answer:
b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e) T = 2π √m / k
h) a = - A w² cos (wt+Ф)
Explanation:
a) see free body diagram in the attachment
b) We write Newton's second law
Fe = m a
-k x = ma
a = -k / m x
c) the acceleration is
a = d²x / dt²
If x = A cos wt
v = dx / dt = -A w sin (wt +Ф)
a = d²x / dt² = - A w² cos (wt+Ф)
d) we substitute in Newton's second law
d²x / dt² = -k / m x
We call
w² = k / m
e) substitute to find w
-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)
w² = k / m
Angular velocity and frequency are related
w = 2π f
f = 1 / T
We substitute
T = 2π / w
T = 2π √m / k
g) v= - A w sin (wt+Ф)
h) acceleration is
a = - A w² cos (wt+Ф)
