During a rockslide, a 660 kg rock slides from rest down a hillside that is 500 m

Question

During a rockslide, a 660 kg rock slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.21. If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the hill, what is the value of U just before the slide? How much energy is transferred to thermal energy during the slide? What is the kinetic energy of the rock as it reaches the bottom of the hill? What is its speed then? m/s

Explanation / Answer

"During a rock-slide, a 600 kg rock slides from rest down a hillside that is 500 m long and 300 m high." Means it is going to gain mgh in energy from falling h = 300 m vertically. "The coefficient of kinetic friction between the rock and the hill surface is 0.23." So some of that energy will be lost to friction, and it won't have all that energy at the bottom. 1. mgh 2. The amount lost to friction is F*d where d = the length of the hill (500 m) and F = the frictional force. So what's F? F = mu*N where mu is the coefficient of friction (given) and N is the normal force. Then what's N? mg*cos(theta) where theta is the angle of the hill. But what's theta you ask? Well, don't happen to know that. But you do know two sides of the triangle, the hypotenuse and the opposite to theta. You can use the Pythagorean theorem to get the other side, then sin(theta) = opp/hyp, which means theta = arcsin(height/length) So that gives you cos(theta), which gives you N, which gives you frictional force, which gives you the work done by friction, which is the "energy transferred to thermal energy". 3. Use the equation => mgh =

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.