A box of weight w=2.0 N accelerates down a rough plane that is inclined at an angle 30 degrees above the horizontal. The normal force acting on the box has a magnitude n= 1.7 N, the coefficient of kinetic friction between the box and the plane is .30, and the displacement d of the box is 1.8 m down the inclined plane.

A)What is the work W done on the box by the weight of the box?

W=m•g•h=m•gvdvsin30

W=m•g•h=m•g•d•sin30

What is the work Wn done on the box by the normal force?

A box with weight w = 900 N is on a rough surface, inclined at an angle of 37 degrees. The box is kept from sliding down (in equilibrium) by means of an external force F. The other forces acting on the box are the normal and friction forces, denoted by n and f. A force diagram, showing the four forces which act on the box, is shown in Figure 4.3. The magnitude of f is 150 N. The external force F is removed and the box accelerates. The magnitudes of the other forces are unchanged. The acceleration of the box is closest to:

To find the work done on the box by the weight of the box, we can use the formula:

Work = Force x Distance x Cos(θ)

where:
- Force is the component of the weight acting down the plane,
- Distance is the displacement of the box down the inclined plane, and
- θ is the angle between the force vector and the displacement vector.

The weight of the box is equal to its mass multiplied by the acceleration due to gravity (W = mg). Assuming we are using standard gravity (9.8 m/s²), the weight of the box can be calculated as:

Weight = 2.0 N

Next, we need to find the component of the weight vector acting down the plane. This can be calculated using trigonometry. The weight vector can be divided into two components: one parallel to the incline (Force parallel) and one perpendicular to the incline (Force perpendicular).

The angle of incline is given as 30 degrees, so we can calculate the component of the weight vector directed down the incline using the formula:

Force parallel = Weight * sin(θ)

Force parallel = 2.0 N * sin(30°)

Force parallel = 2.0 N * 0.5

Force parallel = 1.0 N

Now, we can find the work done by the weight of the box by plugging in the values into the formula:

Work = 1.0 N * 1.8 m * cos(0°)

Work = 1.0 N * 1.8 m * 1.0

Work = 1.8 N·m

Therefore, the work done on the box by the weight of the box is 1.8 joules (J).