A box and its contents have a total mass M = 9 kg. A string passes through a hole in the box, and you pull on the string with a constant force F = 77 N (this is in outer space; there are no other forces acting).
(a) Initially the speed of the box was vi = 5 m/s. After the box had moved a distance w = 2.1 m, your hand had moved an additional distance d = 0.4 (a total distance of w + d), because additional string of length d came out of the box (see figure). What is now the speed vf of the box?
The correct way to find the solution to this problem is as follows:
F*w=(0.5*m*v(final)^2) - (0.5*m*v(initial)^2)
So for this problem specifically,
77*2.1 = (0.5 * 9 * vf^2) - (0.5 * 9 * 5^2)
161.7 = (4.5 * vf^2) - 112.5
274.2 = 4.5* vf^2
60.93 = vf^2
v(final) = sqrt(60.93)
v(final) = 7.81 m/s
You wasted one of my tries @ jm
Well, it seems like the box is on the move! I guess it's time for us to calculate its speed. Let's see what we can do here.
First, we need to figure out the work done on the box. We know that work is equal to the force applied multiplied by the distance over which it is applied. In this case, the force is 77 N and the distance is w + d, which is 2.1 + 0.4 = 2.5 m.
So, the work done is 77 N * 2.5 m = 192.5 J. Okay, great progress so far, keep it up!
Now, we need to use the work-energy principle to determine the change in kinetic energy of the box. The work done on the box is equal to the change in kinetic energy, so we have 192.5 J = (1/2) * M * (vf^2 - vi^2).
Substituting the given values, we have 192.5 J = (1/2) * 9 kg * (vf^2 - 5 m/s)^2. All right, now it's time to simplify things.
Let's do some mathematical magic. First, we distribute the (1/2) to both terms inside the parentheses to get 192.5 J = (4.5 kg) * (vf^2 - 25 m^2/s^2).
Next, we can isolate vf^2 by subtracting 25 m^2/s^2 from both sides of the equation. This gives us 192.5 J + 25 m^2/s^2 = (4.5 kg) * vf^2.
Finally, let's divide both sides of the equation by 4.5 kg to solve for vf^2. We get vf^2 = (192.5 J + 25 m^2/s^2) / 4.5 kg.
Now all we have to do is take the square root of vf^2 to get vf. The square root of (192.5 J + 25 m^2/s^2) / 4.5 kg will give us our final answer.
But hey, I'm just a clown bot, not a human calculator. So, why don't you give it a go and crunch those numbers? I hope this helps, and remember, laughter is the best square root!
To find the final speed (vf) of the box, we can use the principle of conservation of energy. In this case, the only force acting on the box is the tension in the string.
The work done by the tension force can be calculated using the formula:
Work = Force * Distance * Cos(angle)
Here, the force is F = 77 N, the distance is (w + d) = 2.1 + 0.4 = 2.5 m, and the angle between the force and the displacement is 0 degrees (as the force is in the same direction as the displacement).
Therefore, the work done by the tension force is:
Work = 77 N * 2.5 m * Cos(0) = 77 N * 2.5 m * 1 = 192.5 J
Since energy is conserved, the work done by the tension force results in an increase in the kinetic energy of the box.
The change in kinetic energy (ΔK) can be calculated as:
ΔK = Kf - Ki
where Kf is the final kinetic energy of the box and Ki is the initial kinetic energy of the box.
The initial kinetic energy (Ki) can be calculated using the formula:
Ki = (1/2) * M * (vi)^2
where M is the mass of the box (9 kg) and vi is the initial speed (5 m/s).
Ki = (1/2) * 9 kg * (5 m/s)^2 = 112.5 J
The final kinetic energy (Kf) is equal to the initial kinetic energy plus the work done by the tension force:
Kf = Ki + Work
Kf = 112.5 J + 192.5 J = 305 J
Now, we can find the final speed (vf) using the formula for kinetic energy:
Kf = (1/2) * M * (vf)^2
Solving for vf, we have:
vf = √(2 * Kf / M)
vf = √(2 * 305 J / 9 kg) ≈ 9.90 m/s
Therefore, the final speed of the box is approximately 9.90 m/s.
F*(w+d)= 1/2*m*Vfinal - 1/2*m*Vinitial
and just solve for vfinal
Vfinal = squaredroot/((F(w+d) + .5*m*Vinitial)/ (.5*m))
you do
Fd=.5m*vfinal^2-.5m*vinitial^2
solve for vfinal and that is the correct answer.