Question

2. A box of mass 4.00 kg is held in place at the top of 4.00 m long inclined plane. The plane is inclined atan angle of 30.0° and the coefficients of friction between the box and plane are fs = 0.350 and Hk =0.200. The block is released and slides down the plane. At the bottom of the plane is a patch offrictionless ice that is 2.00 m wide. On the other side of the patch of ice is another 4 m long planeinclined at an angle of 30.0°, made from material identical to that of the first plane. The box slides upthis plane, eventually coming to a stop. How far up the second incline does the box slide? Give youranswer to three significant figures.m = 4.00 kg4.00 m?4.00 m30.0°frictionless ice30.002.00 m

Answer 1

The distance the box slides up the second incline can be found by utilizing the conservation of energy principle, taking into account the work done by friction. The initial potential energy minus the energy lost due to friction equals the kinetic energy at the bottom, which then converts back to potential energy as the box climbs the second slope.

Step-by-step explanation:

The question involves the concept of conservation of energy in the context of a classical mechanics problem. The box is released from rest at the top of one incline, slides down, crosses a frictionless patch, and then slides up another incline.

To find out how far up the second incline the box slides, we can use the conservation of energy principle. The potential energy at the top of the first incline is converted into kinetic energy at the bottom, then back into potential energy as it climbs the second incline. However, we must account for energy losses due to friction on the first incline.

The work done by friction (which is the force of friction multiplied by the distance moved along the direction of the force) is subtracted from the initial potential energy to find the kinetic energy at the bottom of the first incline. This kinetic energy, minus any work done by friction on the second incline, gives us the potential energy at the highest point reached on the second incline.

Thus, we can write the equation:

• Initial potential energy - work done by friction on the first incline = kinetic energy at the bottom of the first incline.
• Kinetic energy at the bottom = potential energy at the highest point on the second incline (since the ice is frictionless, there's no loss of energy there).

By solving the equations above, we can find the distance the box slides up the second incline, provided we know the mass of the box, the angle of the incline, the coefficients of friction, and the distances involved.

Answer 2

Step-by-step explanation:

We can represent the situation as follows:

By the Law of conservation of Energy:

`K_i+U_i=K_f+U_f+W_(nc)_{}_{}`

In this case, the box is at rest at the beginning and at the end, so there is no kinetic energy.

Additionally, the Wnc (Work of the non-conservative force) is the work done by the friction.

So, we can rewrite the equation:

`\begin{gathered} U_i=U_f+W_(nc) \\ \text{mgh}_i=\text{mgh}_f+F_f(4+d) \end{gathered}`

Where m is the mass, g is the gravity, hi is the initial height, hf is the final height, Ff is the force of friction in the planes and d is the distance traveled in the second plane.

By the laws of Newton and trigonometry, we can calculate the Friction and hi as follows:

`\begin{gathered} F_f=H_kN=H_kmg\cos 30_{} \\ h_i=4\sin 30=2\text{ m} \end{gathered}`

Additionally, hf is related to d by:

`h_f=d\sin 30`

Now, we can replace this expression on the initial equation to get:

`\begin{gathered} mgh_i=mgd\sin 30+H_kmg\cos 30(4+d_{}) \\ mgh_i=mgd\sin 30+_{}4H_kmg\cos 30+H_kmgd\cos 30 \\ h_i=d\sin 30+4H_k\cos 30+H_kd\cos 30 \end{gathered}`

Finally, we can solve for d and replace the values: