The hill is covered in gravel so that the truck's wheels will slide up the hill instead of rolling up the hill. The coefficient of kinetic friction between the tires and the gravel is k. This design has a spring at the top of the ramp that will help to stop the trucks. This spring is located at height h. The spring will compress until the truck stops, and then a latch will keep the spring from decompressing (stretching back out). The spring can compress a maximum distance x because of the latching mechanism. Your job is to determine how strong the spring must be. In other words, you need to find the spring constant so that a truck of mass mt, moving at an initial speed of v0, will be stopped. For this problem, it is easiest to define the system such that it contains everything: Earth, hill, truck, gravel, spring, etc. In all of the following questions, the initial configuration is the truck moving with a speed of v0 on the level ground, and the final configuration is the truck stopped on the hill with the spring compressed by an amount x. The truck is still in contact with the spring. Solve all of the questions algebraically first. Then use the following values to get a number for the desired answer.

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

The network done is given as;

[tex]\rm W= Fd cos \alpha \\\\ \rm W= F \times D cos 90^0 \\\\\rm W=0\ J[/tex]

Hence the net work done is equal to zero.

(b) The change in the potential energy is equal to the 5.54 ×10⁶ J

[tex]\rm \triangle U_g= mg (h+x sin\theta)\\\\ \rm \triangle U=1200 \times 9.81(45+3.5sin 37.8^0)\\\\ \rm \triangle U_g=5.5443 \times 10^6 \ J[/tex]

Hence the changes in the potential energy are equal to the 5.54 ×10⁶ J.

(c) The change in the thermal energy will be 3.3425×10⁶ J.

The formula for the thermal energy change is found as;

[tex]\rm \triangle E_{thermal} =\mu_k mgcos \theta (\frac{h-L}{sin \theta} +x )\\\\ \rm \triangle E_{thermal} =0.6 \times 1200 \times 9.81 \times 37.8^0 ( (\frac{45-18.4}{sin 37.8^0} +3.5 )\\\\ \rm \triangle E_{thermal} =3.342 \times 10^6 \ J[/tex]

Hence the changes in the thermal energy will be 3.3425×10⁶ J.

To learn more about the work done refer to the link ;

https://brainly.com/question/3902440