Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
Answer:
Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Answer:
Explanation:I don't say you must have to mark my ans as brainliest but if you think it has really helped u plz don't forget to thank me....
It traveled 200 m in 50 seconds. 200/50 can be simplified to 4 m/s!
The velocity is -4 m/s (negative because it travelled from 100 to -100 or backwards)
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