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: A Neutron.
I know this is correct. Thank's and yw :3
Answer:
It requires more tension to pull up the track
Explanation:
Net force must be zero to maintain constant velocity.
Weight force will always be pointed down the slope. Call it W
Friction force (Call it Ff) will be down slope when movement is up slope.
Friction force will be up slope when movement is down slope.
W and Ff are always positive numbers
Call the pulling force T
If Up slope is considered the positive direction
Moving up slope
Tu - Ff - W = 0
Tu = Ff + W
Moving down slope
Td + W - Ff = 0
Td = Ff - W
Ff + W > Ff - W therefore Tu > Td
Answer:
a ) = 381.48 J
b )= 84.25 cm
Explanation:
Kinetic energy of the runner
= 1/2 m v²
= .5 x 66 x 3.4²
= 381.48 J
The final kinetic energy of the runner is zero .
Loss of mechanical energy
= 381.48 J
This loss in mechanical energy is due to action of frictional force .
b )
Let s be the distance of slide
deceleration due to frictional force
= μmg/m
.7 x 66 x 9.8 / 66
a = - 6.86 m s⁻¹
v² = u² - 2 a s
0 = 3.4² - 2x6.86 s
s = 3.4² / 2x6.86
= .8425 m
84.25 cm
More force needs to be applied
