Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 =
let's calculate
v₁ =
v₁ = 37.5 cm / s
Answer:
20.4
Explanation:
To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.
First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.
Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.
Applied force = 1055 N (490 + 565)
Friction force= Unknown
To find the friction force, use the kinetic friction formula, Friction = μkN
μk is the coefficient, which the problem includes- it is 0.613.
N is the normal force, which we have to find.
*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.
Weight = mg = (40)(9.8) = 392 N
Normal force = 392 N
Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N
Friction = 240.296 N
Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N
Horizontal Net Force: 814.704 N
Now that we know the net force, plug in the numbers for the formula
∑F= ma.
814.704 = (40.0)(a)
*Divide on both sides)
a = 20.3676 m/s^2
Round it to 3 significant figures, to get:
20.4
Answer:
a) Fg = -27.4 N
b) Fnet = 5.2 N
c) a = 1.9 m/s2
Explanation:
a)
b)
c)