9) A force in the +x-direction with magnitude F(x) = 18.0 N – (0.530 N/m) x is applied to a 6.00-kg box that is sitting on a frictionless surface. F(x) is the only horizontal force on the box. If the box is initially at rest at x = 0, what is its speed after it has traveled 14.0 m?
Take the integral of F(x) which is
18.0x-.5(.530)x^2 then plug in 14.0 for x
=200.06J
so .5mv^2=W since v(initial) is zero so then solve for v
v(final)=((2*200.06)/6)^(1/2)
v(final)=8.17m/s
this is wrong
when you integrate x you get (x^2)/2, the .5 represents the 1/2
Sir mirian. May I ask why is their a .5 next to (0.53)?
Well, let's see. The force applied to the box in the +x-direction is given by F(x) = 18.0 N – (0.530 N/m) x. To find the speed of the box after it has traveled 14.0 m, we need to calculate the work done on the box by this force.
The work done on an object is given by the formula W = ∆K, where W is the work done, ∆K is the change in kinetic energy, and K is the kinetic energy.
Since the box is initially at rest, its initial kinetic energy is zero. So, the work done on the box is equal to its final kinetic energy.
The work done by a variable force is given by the integral of the force applied over the distance traveled. So, let's set up the integral:
W = ∫[0,14.0] (18.0 N – (0.530 N/m) x) dx
Now, we can solve this integral to find the work done on the box.
But, wait! Before we continue, let me tell you a joke to lighten up the mood:
Why don't scientists trust atoms?
Because they make up everything!
Okay, back to the problem. Let's solve the integral:
W = ∫[0,14.0] (18.0 N – (0.530 N/m) x) dx
= [18.0 N * x - (0.530 N/m) * x^2 / 2] evaluated from 0 to 14.0
= (18.0 N * 14.0) - (0.530 N/m) * (14.0^2 / 2)
Now that we have the work done, we can set it equal to the final kinetic energy of the box to find its speed.
W = ∆K
(18.0 N * 14.0) - (0.530 N/m) * (14.0^2 / 2) = (1/2) * m * v^2
Let's plug in the values and solve for v.
Don't worry, I've got a joke to make the wait more enjoyable:
Why don't skeletons fight each other?
Because they don't have the guts!
Okay, now we can solve it:
(18.0 N * 14.0) - (0.530 N/m) * (14.0^2 / 2) = (1/2) * 6.00 kg * v^2
Once you've calculated the final speed, you'll have your answer!
To determine the speed of the box after it has traveled 14.0 m, we need to apply the concept of work and energy.
First, let's find the net work done on the box by the applied force during the displacement of 14.0 m.
The work done (W) is given by the equation:
W = F * d * cosθ
In this case, the applied force F(x) is given as F(x) = 18.0 N – (0.530 N/m) x. The displacement d is 14.0 m. Since the force is applied in the positive x-direction, the angle θ between the force and displacement vectors is 0 degrees (cosine of 0 degrees is 1).
Therefore, the net work done is:
W = F * d * cosθ
W = (18.0 N – (0.530 N/m) * 14.0 m) * cos(0°)
Now, let's calculate the work done:
W = (18.0 N – 7.42 N) * 14.0 m
W = 10.58 N * 14.0 m
W = 148.12 J
Next, we can use the work-energy theorem to find the change in kinetic energy of the box:
ΔK = Kf - Ki
Where ΔK is the change in kinetic energy, Kf is the final kinetic energy, and Ki is the initial kinetic energy.
Since the box is initially at rest, the initial kinetic energy is zero (Ki = 0).
Therefore, ΔK = Kf - 0
ΔK = Kf
The work done on the box is equal to the change in kinetic energy:
W = ΔK
148.12 J = Kf
Finally, we can use the equation for kinetic energy to find the speed (v) of the box:
Kf = 1/2 * m * v^2
Substituting the known values:
148.12 J = 1/2 * 6.00 kg * v^2
Now, solve for v:
2 * (148.12 J) = 6.00 kg * v^2
296.24 J = 6.00 kg * v^2
v^2 = 296.24 J / 6.00 kg
v^2 = 49.374 m^2/s^2
Taking the square root of both sides:
v = √(49.374 m^2/s^2)
v ≈ 7.03 m/s
Therefore, the speed of the box after it has traveled 14.0 m is approximately 7.03 m/s.
Fx = 18-0.53N/m * 14m = 10.58 N.
a = F/m = 10.58/6 = 1.76 m/s^2.
V^2 = Vo^2 + 2a*d = 0 + 2*1.76*14 = 49.4
V = 7.03 m/s.