A painter lifts a 2.8 kg bucket using a rope. if the acceleration of the bucket in 3.1 m/s2, what is the tension in the rope (in
n)? selected:
a. 27.4 n
b. 9.8 n
c. 8.68 n
d. 36.1 n
e. 2.94 n f. 0 n
a. 27.4 n
b. 9.8 n
c. 8.68 n
d. 36.1 n
e. 2.94 n f. 0 n
2 answers:
Refer to the diagram shown below
Given:
m = 2.8 kg, the mass of the bucket
a =3.1 m/s², the acceleration of the bucket
Therefore
W = 2.8*9.8 = 27.44 N, the weight of the bucket.
Let T = the tension in the rope.
From the free body diagram, the net force accelerating the bucket is
T - W = m*a
That is,
T = W + m*a
= 27.44 + 2.8*3.1 N
= 36.12 N
Answer: d. 36.1 N
Given:
m = 2.8 kg, the mass of the bucket
a =3.1 m/s², the acceleration of the bucket
Therefore
W = 2.8*9.8 = 27.44 N, the weight of the bucket.
Let T = the tension in the rope.
From the free body diagram, the net force accelerating the bucket is
T - W = m*a
That is,
T = W + m*a
= 27.44 + 2.8*3.1 N
= 36.12 N
Answer: d. 36.1 N
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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)
