Physics 201 Homework 5


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1 Physics 201 Homework 5 Feb 6, The (nonconservative) force propelling a 1500kilogram car up a mountain joules road does joules of work on the car. The car starts from rest at sea level and has a speed of 27.0 m/s at an altitude of 200 meters above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces. The propelling force has the effect of increasing the energy level of the car. This energy goes three places: kinetic energy (or speed), potential energy (or altitude), and some is lost to the nonconservative friction. We know the energy going in and we can calculate the amount that goes into kinetic and potential energies. What is left must have been lost to the nonconservative forces. In symbols: We know the kinetic energy gained: And the potential energy gained: W added W lost = E = KE + P E KE = 1 2 mv2 = 1 2 (1500)(27.0)2 = P E = mgh = (1500)(9.80)(200) = Plugging these and the energy added we have: ( ) W lost = = W lost = It s a technicality, but the question asks for the work done by the resistive forces. Since we have calculated the work lost, we need to flip the sign to emphasize that the energy level is decreasing as a consequence of the resistance. 2. A person pushes a 16.0kilogram shopping cart at a constant velocity for a (a) 54.9 newtons distance of 22.0 meters. She pushes in a direction 29.0 below the horizontal. (b) 1060 joules (c) joules A 48.0newton frictional force opposes the motion of the cart. (a) What is the (d) none magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force. (a) Since the cart is moving at constant velocity, the system is in equilibrium, so the net force is zero. In this problem there are four forces involved. The pushing force, the weight of the cart, the normal support force, and the kinetic friction. The motion is horizontal, so we should align the coordinate that way. The components of the net force are: F net,x = P x F F net,y = P y W + N Both of these components sum to zero. Since we are given the friction force (48.0 newtons), we know that the xcomponent of the pushing force is also 48.0 newtons. Since we know that the canonical angle of the push is 331, we have: P x = P cos 331 = 48.0 = P =
2 (b) The work done by any force is the product of the displacement and the component of the force parallel to the displacement. In this case, x = 22.0 meters and P x = 48.0 newtons. Thus, W = xp x = (22.0)(48.0) = 1056 (c) Similarly for the friction, but this force opposes the motion so its component is negative: W = xf x = (22.0)( 48.0) = 1056 (d) The gravitational force has no component in the x direction. Thus, W = xw x = (22.0)(0) = 0 3. A kilogram arrow is fired horizontally. The bowstring exerts an average 39 m/s force of 65 newtons on the arrow over a distance of 0.90 meters. With what speed does the arrow leave the bow? The work done by the bowstring is W = xf x = (0.90)(65) = 58.5 This work has the effect of increasing the energy of the arrow. In this case it increases the kinetic energy of the arrow. The initial kinetic energy is zero, so KE f = 58.5 Using the formula for kinetic energy, KE = 1 2 mv2, we can derive the final resulting speed: 1 2 (0.075)(v)2 = 58.5 = v = An extreme skier, starting from rest, coasts down a mountain slope that makes 10.9 m/s an angle of 25.0 with the horizontal. The coefficient of kinetic friction between her skis and the snow is She coasts down a distance of 10.4 meters before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.50 meters below the edge. How fast is she going just before she lands? There are two phases to the motion in this problem. The first coasting phase, is basically an inclined plane problem from Chapter 4. At the end of this phase the skier will have a certain velocity angled down 25 from the horizontal. The second phase is standard projectile problem which we could solve using the techniques from Chapter 3. However, this since we are only asked about the speed, we could use kinetic energy to answer the question. Either approach will work. But since this is a Chapter 6 problem, it seems appropriate to use energy. In the first phase, there are three forces at work: weight, support, and friction. Using coordinates aligned with the motion (assume the slope is to the right), the canonical angles associated with these forces are 295, 90, and 180, respectively. The magnitudes of these forces are W = mg, N, and F k = (0.200)(N), respectively. We don t know m, but we will assume it cancels out. We don t know N, but it must counterbalance the ycomponent of the weight. Thus, N + W y = 0. Therefore, N = W y = W sin θ = (9.80)(m) sin 295 = (8.8818)(m) Now there is motion in the xdirection. According to Newton s second law we have F net = ma. The net force is in the xdirection, but friction points in the negative direction, so F net,x = W x F k = (m)(9.80) cos 295 (0.200)(8.8818)(m) = (2.3653)(m) 2
3 So, the acceleration of the skier down the slope is m/s 2. Now we have a constant acceleration problem. The quantities are a = v 0 = 0 v =? x = 10.4 The equation to use is v 2 = v ax. Thus, (v) 2 = (0) 2 + (2)(2.3653)(10.4) = v = This is the endpoint of the first phase of motion, but it is also the beginning of the second phase. We decided to use energy considerations to work this phase. We know that she is 3.50 meters above her landing point. This means she has a certain amout of potential energy: And her initial kinetic energy is P E = mgh = (m)(9.80)(3.50) = (34.300)(m) KE = 1 2 mv2 = 1 2 (m)(7.0141)2 = (24.599)(m) Since no energy is lost, the potential energy is converted into additional kinetic energy at the end of the drop. So the final kinetic energy is KE = (34.300)(m) + (24.599)(m) = (58.899)(m) Using the formula for kinetic energy, KE = 1 2 mv2, we can determine her final speed: (58.899)(m) = 1 2 (m)(v)2 = v = Rocket man has a propulsion unit strapped to his back. He starts from rest 23 kj on the ground, fires the unit, and is propelled straight upward. At a height of 16 meters, his speed is 5.0 m/s. His mass, including the propulsion unit, has the approximately constant value of 136 kilograms. Find the work done by the force generated by the propulsion unit. The gravitational force is conservative, so we can say that W = E = KE + P E Now, initially Rocket man has no kinetic or potential energy because he is at rest and on the ground. But at the end point, he has both kinetic and potential energy. Thus, KE = 1 2 mv2 = 1 2 (136)(5.0)2 = 1700 and P E = mgh = (136)(9.80)(16) = Now, his initial mechanical energy is zero. But the final mechanical energy is E = KE + P E = = The work done by the propulsion unit is where this energy came from. 6. In 2.0 minutes, a ski lift raises four skiers at constant speed to a height of 3000 watts 140 meters. The average mass of each skier is 65 kilograms. What is the average power provided by the tension in the cable pulling the lift? 3
4 Since the work done by the lift has the effect of increasing the potential energy of the skiers, we know by the workenergy theorem that W = P E = mgh = (4 65)(9.80)(140) = Since this quantity of work is done over a 2.0 minute time period (or 120 seconds), by definition, the power involved is P = W ork t = = A small lead ball, attached to a 1.5 meter rope, is being whirled in a circle 42 meters that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 0.75 meters above the ground. The fall travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise? At the top of its trajectory, the kinetic energy is momentarily zero all the energy is in the gravitational potential energy mgh. From this we can calculate the height. Since we know the initial height, all we need to determine is the initial speed. The ball moves in uniform circular motion with a radius of 1.5 meters. Each revolution is 2πr = meters. The speed is three times this value per second: So, the energy of the ball is v = (3)(2π)(1.5) = KE = 1 2 mv2 = 1 2 (m)(28.274)2 = PE = mgh = (m)(9.8)(0.75) = 7.35 For a total of (407.07)(m) joules of energy. Therefore the maximum height must be PE = mgh = (m)(9.8)(h) = (407.07)(m) = h = A pendulum consists of a small object hanging from the ceiling at the end 43 of a string of negligible mass. The string has a length of 0.75 meters. With the string hanging vertically, the object is given an initial velocity of 2.0 m/s parallel to the ground and swings upward on a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle θ with its initial vertical orientation and then swings back downward. Find the angle θ. Initially the height of the pendulum is zero, so its potential energy is zero. The kinetic energy is KE = 1 2 mv2 = 1 2 (m)(2.0)2 = (2.0)(m) We can use this to determine the height the pendulum rises, since the energy will all be converted to potential: PE = mgh = (m)(9.8)(h) = (2.0)(m) = h = From this we should be able to calculate the angle in question using some trig. I will help to draw a little picture here: 4
5 0.75 θ Figure 1: Problem 6.46 From the picture we can see the triangle defined by the angle in which we are interested. We know the hypotenuse is 0.75 meters and the adjacent side is ( ) = These sides are connected via the cosine function. Thus: cos θ = = θ = A semitrailer is coasting downhill along a mountain highway when its brakes 27.0 m/s fail. The driver pulls onto a runawaytruck ramp that is inclined at an angle of 14.0 above the horizontal. The semitrailer coasts to a stop after traveling 154 meters along the ramp. What was the truck s initial speed? Neglect air resistance and friction We are asked about the initial state, so let s start with the end state instead. We know the semi comes to a stop, so there is no kinetic energy. It is at a certain height, which is potential energy. But how high? The height is actually the opposite side of the triangle formed by the ramp. The hypotenuse is 154 meters and the angle is Thus: sin 14.0 = h 154 = h= So, the potential energy is PE = mgh = (m)(9.8)(37.256) = (365.11)(m) This is how much kinetic energy it has when it hits the ramp. Thus, (365.11)(m) = 1 2 mv2 = v = A 1300kilogram car is to accelerate from rest to a speed of 30.0 m/s in a 132 horsepower time of 12.0 seconds as it climbs a 15 hill. Assuming uniform acceleration, what minimum horsepower is needed to accelerate the car in this way? We could use the techniques from Chapter 4 to solve this, but using energy is much easier. The initial energy of the car is zero: it starts at rest (no kinetic) and is at the bottom of the hill (no potential). As the engine works on the car, it gains energy. At the top of the hill it has kinetic energy: KE = 1 2 mv2 = 1 2 (1300)(30.0)2 = Its potential energy is governed by it height. In order to figure this out we need to determine how far it travels. We can use x = 1 2 (v + v 0)(t) = 1 2 ( )(12.0) = 180 5
6 Since the hill is at a 15 incline, the total elevation change is So the car s final potential energy is h = x sin θ = (180) sin 15 = P E = mgh = (1300)(9.8)(46.587) = The work done by the engine is the combination of the kinetic and potential energies: W = Finally, power is defined as work done per second, so: P = W t = = Since there are 746 watts in one horsepower, we have P = 98210/746 = hp 6
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