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

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:

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