Answer:
1.5 m/s²
Explanation:
For the block to move, it must first overcome the static friction.
Fs = N μs
Fs = (45 N) (0.42)
Fs = 18.9 N
This is less than the 36 N applied, so the block will move. Since the block is moving, kinetic friction takes over. To find the block's acceleration, use Newton's second law:
∑F = ma
F − N μk = ma
36 N − (45 N) (0.65) = (45 N / 9.8 m/s²) a
6.75 N = 4.59 kg a
a = 1.47 m/s²
Rounded to two significant figures, the block's acceleration is 1.5 m/s².
Usually the coefficient of static friction is greater than the coefficient of kinetic friction. You might want to double check the problem statement, just to be sure.
I'm not sure about the magnitude, but the direction of the normal force is upward.
1)a 2)D 3)a. I think the answers are
Answer:
a) Yes
b) 7 rad/s
c) 0.01034 J
Explanation:
a)
Yes the angular momentum of the block is conserved since the net torque on the block is zero.
b)
m = mass of the block = 0.0250 kg
w₀ = initial angular speed before puling the cord = 1.75 rad/s
r₀ = initial radius before puling the cord = 0.3 m
w = final angular speed after puling the cord = ?
r = final radius after puling the cord = 0.15 m
Using conservation of angular momentum
m r₀² w₀ = m r² w
r₀² w₀ = r² w
(0.3)² (1.75) = (0.15)² w
w = 7 rad/s
c)
Change in kinetic energy is given as
ΔKE = (0.5) (m r² w² - m r₀² w₀²)
ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)
ΔKE = 0.01034 J
