Question:

A 2 kg box is initially at rest on a frictionless horizontal surface. A 10 N constant horizontal force is applied to the box, moving it a distance of 5 meters in the positive x-direction.

a) Calculate the work done by the applied force on the box.

b) Calculate the change in kinetic energy of the box.

c) Calculate the final speed of the box.

(Note: Take the positive x-direction as the direction of the applied force and motion.)

Answer:

a) The work done by a constant force can be calculated using the equation:

Work = Force ⨉ Distance ⨉ cosθ

Where θ is the angle between the force vector and the displacement vector.

In this case, the force and displacement vectors are in the same direction, so the angle θ is 0 degrees. Therefore, the cosθ term is equal to 1.

Given Force = 10 N and Distance = 5 m, we can substitute these values into the equation:

Work = 10 N ⨉ 5 m ⨉ cos(0°) = 10 N ⨉ 5 m ⨉ 1 = 50 Joules

Therefore, the work done by the applied force on the box is 50 Joules.

b) The change in kinetic energy (ΔKE) of an object can be calculated using the equation:

ΔKE = Work

Therefore, the change in kinetic energy of the box is also 50 Joules.

c) The final speed of the box can be calculated using the equation:

ΔKE = (1/2)mv^2

Where ΔKE is the change in kinetic energy, m is the mass of the object, and v is the final velocity.

We already know the change in kinetic energy (ΔKE = 50 Joules) and the mass of the box (m = 2 kg).

Substituting these values into the equation, we can solve for the final velocity (v):

50 Joules = (1/2)(2 kg)v^2

v^2 = (50 Joules) / (1 kg) = 50 m^2/s^2

Taking the square root of both sides to solve for v:

v = √(50 m^2/s^2) ≈ 7.07 m/s

Therefore, the final speed of the box is approximately 7.07 m/s.