A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.30 kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.

Part A
If the box is initially at rest at x=0, what is its speed after it has traveled 16.0 m ?
Express your answer to three significant figures and include the appropriate units.

We know that work = ʃ F∙dx = 18x - 0.53x²/2 evaluated from 0 to 16

= 18*16 - 0.53*16²/2 = ......... of work done on the box
Therefore the KE of the box = above work = ½*m*V²
-----> V = sqrt(2* work)

Why did the box go to see the doctor?

Because it had a case of velocity!

To find the speed of the box after it has traveled 16.0 m, we can use the work-energy theorem. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.

The work done on the box can be calculated by integrating the force function over the distance traveled. The force function is given as F(x) = 18.0N - (0.530N/m)x. The distance traveled is 16.0 m.

The work done on the box is given by:

W = ∫(F(x)dx) from x=0 to x=16.0

W = ∫((18.0N - (0.530N/m)x)dx) from x=0 to x=16.0

W = [18.0x - (0.530N/m)x^2/2] from x=0 to x=16.0

W = [18.0(16.0) - (0.530N/m)(16.0)^2/2] - [18.0(0) - (0.530N/m)(0)^2/2]

W = 288.0 - (0.530N/m)(128.0)

W = 288.0 - 67.84N

The work done on the box is 220.16 J (joules).

According to the work-energy theorem, the work done on the box is equal to the change in kinetic energy:

W = ΔKE

220.16 J = (1/2)mv^2 - (1/2)mv_initial^2

Here, m = 8.30 kg is the mass of the box, and v_initial is the initial speed of the box, which is zero since it is initially at rest.

220.16 J = (1/2)(8.30 kg)v^2 - (1/2)(8.30 kg)(0)^2

220.16 J = 4.15v^2

v^2 = 220.16 J / 4.15 kg

v^2 = 53.05735 m^2/s^2

Taking the square root of both sides, we get:

v = sqrt(53.05735) m/s

v ≈ 7.280 m/s

Therefore, the speed of the box after it has traveled 16.0 m is approximately 7.280 m/s.

To find the speed of the box after it has traveled 16.0 m, we need to use the principle of work and energy. The work done on an object is given by the equation:

Work = ΔKinetic Energy

The work done by a force is the product of the force and the displacement:

Work = Force × Displacement × cosθ

In this case, the force is given by F(x) = 18.0N − (0.530N/m)x, and the displacement is 16.0 m. Since the force is in the +x direction and the displacement is also in the +x direction, the angle θ between the force and the displacement is 0 degrees, so cosθ = 1.

Let's calculate the work done by the force on the box:

Work = (18.0N − (0.530N/m)x) × 16.0 m × cos(0°)

Simplifying the equation:

Work = (18.0N − (0.530N/m) × 16.0 m)

Now, let's find the change in kinetic energy of the box. The initial kinetic energy (K1) is zero since the box is initially at rest. The final kinetic energy (K2) is the kinetic energy of the box after it has traveled 16.0 m.

ΔKinetic Energy = K2 − K1

Since K1 = 0, we can simplify it further:

ΔKinetic Energy = K2

Now we can equate the work done by the force to the change in kinetic energy:

(18.0N − (0.530N/m) × 16.0 m) = K2

Now let's solve for K2:

K2 = 18.0N × 16.0 m − (0.530N/m) × 16.0 m²

K2 = 288.0 N·m − 8.48 N·m

K2 = 279.52 N·m

The final step is to find the speed of the box. The formula for kinetic energy is:

Kinetic Energy = (1/2) × mass × velocity²

Since we know the mass of the box is 8.30 kg, we can rearrange the equation to solve for the velocity:

velocity = sqrt((2 × Kinetic Energy) / mass)

Plugging in the values:

velocity = sqrt((2 × 279.52 N·m) / 8.30 kg)

velocity = sqrt(67.28 N·m/kg)

Now, let's calculate the velocity:

velocity ≈ 8.20 m/s

Therefore, the speed of the box after it has traveled 16.0 m is approximately 8.20 m/s.