The given question is incomplete. The complete question is as follows.
The block has a weight of 75 lb and rests on the floor for which = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
Determine the output of the motor at the instant .
Explanation:
We will consider that equilibrium condition in vertical direction is as follows.
N - W = 0
N = W
or, N = 75 lb
Again, equilibrium condition in the vertical direction is as follows.
= 0
=
= 30 lb
Now, the equilibrium equation in the horizontal direction is as follows.
or, T =
=
=
= 17.32 lb
Now, we will calculate the output power of the motor as follows.
P = Tv
=
=
= 0.189 hp
or, = 0.2 hp
Thus, we can conclude that output of the given motor is 0.2 hp.
There is no correct description on that list of choices.
Answer:
Explanation:
easy way
when system is all kinetic energy, velocity is at a maximum
E = ½mv²
v = √(2E/m) = √(2(25)/0.5) = √100 = 10 m/s
harder way
ω = √(k/m) = √(80/0.5) = √160 rad/s
When the system is entirely spring potential, the amplitude A is
E = ½kA²
A = √(2E/k) = √(2(25)/80) = 0.790569... = 0.79 m
maximum velocity is ωΑ = 0.79√160 = 10 m/s
OK, the wedge is accelerating (a) at Theta = 180 degrees (to the right) and the wedge is inclined theta = 75 degrees. For the m = 2 kg block to remain at rest all we need is a net force f = W cos(theta) - F sin(theta) = 0; where F = ma and W = mg the weight of the block. That is, the weight component along the incline is offset by the acceleration component along the surface; so the block does not slide.
Solving we have W cos(theta) = mg cos(theta) = ma sin(theta) = F sin(theta); such that a = g cos(theta)/sin(theta) = g cot(theta). Assuming g ~ 9.81 m/sec^2, you can now plug and chug to find the answer.
<span>The physics is this...when the net force on a body is f = 0, that body will not accelerate and start to move if it is already still. So when the block's weight component along the surface of the wedge is offset by the equal but opposite force along the surface of the accelerating wedge, the still block will not move.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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