Transcript 1.78 m/s² = a - Cloudfront.net

(Mon) A 3.00 kg block starts from rest at the top of a
30.0º incline and accelerates uniformly down the incline
moving 2.00 m in 1.50 s. What is the magnitude of the
acceleration of the block? (5 min / 5 pts)
(Mon) A 3.00 kg block starts from rest at the top of a
30.0º incline and accelerates uniformly down the incline
moving 2.00 m in 1.50 s. What is the magnitude of the
acceleration of the block? (5 min / 5 pts)
x Dir
y Dir
Δx
2.0 m
N/A
vi
0 m/s
N/A
N/A
vf
a
?????
N/A
Δx = vit + ½ at²
t
1.5 s
N/A
2.0 m = 0 + ½ a(1.5s)²
2.0 m / (0.5)(2.25 s²) = a
1.78 m/s² = a
(Tue) Veteran’s Day – No School
(Wed) A 3.00 kg block starts from rest at the top of a
30.0º incline and accelerates uniformly down the incline
moving 2.00 m in 1.50 s. If the acceleration down the
block is 1.78 m/s², what is the magnitude for the net
force for the block down the ramp? (5 min / 5 pts)
(Wed) A 3.00 kg block starts from rest at the top of a
30.0º incline and accelerates uniformly down the incline
moving 2.00 m in 1.50 s. If the acceleration down the
block is 1.78 m/s², what is the magnitude of the net force
for the block down the ramp? (5 min / 5 pts)
F = ma
F = (3.00 kg) x (1.78 m/s²)
F = 5.34 kgm/s²
F = 5.34 N
(Thu) A 3.00 kg block starts from rest at the top of a
30.0º incline and accelerates uniformly down the incline
moving 2.00 m in 1.50 s. If the force due to friction is
-9.37 N and the net force down the ramp is 5.34 N, what
is the coefficient of friction between the block and the
ramp? (8 min / 5 pts)
(Thu) A 3.00 kg block starts from rest at the top of a
30.0º incline and accelerates uniformly down the incline
moving 2.00 m in 1.50 s. If the force due to friction is
-9.37 N and the net force down the ramp is 5.34 N, what
is the coefficient of friction between the block and the
ramp? (8 min / 5 pts)
x Dir
y Dir
In the ‘y’ direction
Fnet = Fg + Fn + Ff
Fg
-25.49 N
Fn
0.0 N
Ff
-9.37 N
0.0 N
Fnet
5.34 N
0.0 N
25.49 N
Fg,y = magcos(30)
Fg,y = (3.00kg)(9.81 m/s²)(0.86)
Fg,y = 25.49 N
0 N = -25.49 N + Fn + 0 N
25.49 N = Fn
Ff = (Fn)(μs)
(μs) = 0.37
Ff/(Fn) = (μs)
(μs) = 9.37 N / 25.49 N
(Fri) A 3.00 kg block starts from rest at the top of a 30.0º
incline and accelerates uniformly at a rate of 1.78 m/s²
down the incline moving 2.00 m in 1.50 s. What is the
magnitude of the velocity for the block at the bottom
after traveling the 2.00 m? (5 min / 5 pts)
(Fri) A 3.00 kg block starts from rest at the top of a 30.0º
incline and accelerates uniformly at a rate of 1.78 m/s²
down the incline moving 2.00 m in 1.50 s. What is the
magnitude of the velocity for the block at the bottom
after traveling the 2.00 m? (5 min / 5 pts)
x Dir
y
Dir
N/A
vi
2.0 m
0 m/s
vf
?????
N/A
Δx
a
t
N/A
1.78 m/s² N/A
1.5 s
N/A
vf² = vi² + 2aΔx
vf² = 0² + 2(1.78 m/s²)(2.0 s)
vf² = 7.12 m²/s²
vf = 2.67 m/s
End of Week Procedures:
1. Add up all the points you got this
week
2. Put the total number of points at the
top of your page (out of 20 points)
3. List any dates you were absent (and
why)
4. Make sure your name and period is
on the top of the paper
5. Turn in your papers