A 15 kg box sits still on a rough inclined plane (no math required here).
1 answer:
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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)
Answer:
D. 1.8 × 102 newtons radially inward
Explanation:
The magnitude of the centripetal force is given by:
where
m is the mass of the object
v is the tangential speed
r is the radius of the circular trajector
In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get
The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is
D. 1.8 × 102 newtons radially inward