Worksheet #06 1. James Bond (90.0 kg), outfitted with perfectly matching skis and skiware, is at the top of a steep slope that a secret spy like him can easily handle. He lets himself go from rest and smoothly slides down the 15.0 m-high hill. A big parking lot lies at the bottom of the hill. Since the parking area has been cleared of snow, the friction between the ground and the skis brings our hero to a halt at point D, located at a distance d = 12.0 m from point C. The descent can be considered frictionless. Take the potential energy to be zero at the bottom of the slope. (a) What is the mechanical energy of James Bond at points A and D? (b) Determine the speed of Bond at position B abd C. (c) What is the work done by friction in the parking lot? (d) Find the magnitude of the average friction force in part (1c). 1
Worksheet #06 2. An object of mass m = 3.00 kg is released from rest at a height of h = 5.00 m on an inclined ramp which makes an angle θ = 10.0 with the horizontal. At the foot of the ramp and after a horizontal surface of length d = 13.0 m is the tip of a spring of force constant k = 4000 N/m (see figure). The object slides down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. Find x and describe what happens to the object after it comes to rest (a) if all surfaces are frictionless. (b) If the coefficient of kinetic friction µ k between the horizontal surface and the object is 0.200. 2
Worksheet #06 3. A particle of mass m = 2.0 kg moves in a region of space with the following potential energy function. 60 40 20 P E(J) 0. 20 40 60 0 2 4 6 x(m) 8 10 12 (a) Find the work done on the particle when the particle moves from x = 4 m to x = 5 m. (b) Assume the particle is initially at x = 11 m and has a velocity v 0 = 2 5 m/s in the positive x direction. Find the total mechanical energy of the particle. (c) Where is the speed of the particle minimum and maximum? 3
Worksheet #06 (d) Describe qualitatively the motion of the particle. (e) Now imagine that the total mechanical energy of the particle is E = 40 J. Describe the motion of the particle. Where is the speed of the particle maximum/minimum? Determine these speeds. 4
Solution for Worksheet #06 1. James Bond (90.0 kg), outfitted with perfectly matching skis and skiware, is at the top of a steep slope that a secret spy like him can easily handle. He lets himself go from rest and smoothly slides down the 15.0 m-high hill. A big parking lot lies at the bottom of the hill. Since the parking area has been cleared of snow, the friction between the ground and the skis brings our hero to a halt at point D, located at a distance d = 12.0 m from point C. The descent can be considered frictionless. Take the potential energy to be zero at the bottom of the slope. (a) What is the mechanical energy of James Bond at points A and D? Answer: E A = mgh A = (90.0 kg) ( 9.80 m/s 2) (15.0 m) = 13 200 J (1) E D = 0 At rest; At zero height! (2) (b) Determine the speed of Bond at position B abd C. Answer: Using conservation of energy: E A = E B mgh = mg h 2 + 1 2 mv2 B v B = E A = E C mgh = 1 2 mv2 C v C = gh = (9.80 m/s 2 ) (15.0 m) = 12.1 m/s (3) 2gh = 2 ( 9.80 m/s 2) (15.0 m) = 17.1 m/s (4) (c) What is the work done by friction in the parking lot? Answer: Using conservation of energy (where the difference in energy between points A and D is what is lost due to friction along the path.) E D E A = W f C D W f C D = 0 13 200 J = 13 200 J (5) (d) Find the magnitude of the average friction force in part (1c). Answer: From the definition of work: C D W f = f k d f k = W f C D 13 200 J = d 12.0 m = 1100 N (6) 1
Solution for Worksheet #06 2. An object of mass m = 3.00 kg is released from rest at a height of h = 5.00 m on an inclined ramp which makes an angle θ = 10.0 with the horizontal. At the foot of the ramp and after a horizontal surface of length d = 13.0 m is the tip of a spring of force constant k = 4000 N/m (see figure). The object slides down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. Find x and describe what happens to the object after it comes to rest (a) if all surfaces are frictionless. Answer: Initially, all energy is gravitational potential energy. When the object fully compresses the spring at the bottom of the incline, all the energy stored in the spring as potential energy (the object comes momentarily to rest). So E initial = E final mgh = 1 2mgh 2 kx2 x = = 2 (3.00 kg) (9.80 m/s2 ) (5.00 m) = 0.271 m (7) k 4000 N/m After the object has completely compressed the spring, the spring will start to convert its potential energy back to kinetic (pushing the object). The object then leaves the spring with maximum kinetic energy 1 2 mv2 = 1 2 kx2. Then the object will climb the incline back reaching its original height! (b) If the coefficient of kinetic friction µ k between the horizontal surface and the object is 0.200. Answer: When friction is present, the final energy will be less than initial energy by the magnitude of work lost to friction. When the spring is compressed by a distance x, the object has traveled on the horizontal surface for a a distance of (d + x). Using the conservation of energy theorem: E initial + W f = E final mgh µ k mg (d + x) = 1 2 kx2 x 2 + 2µ kmg x 2mg k k (h µ kd) = 0 (8) x 2 + (0.002 94 m) x 0.0353 m 2 = 0 (9) The last equation is quadratic in x. The positive root is: x = 0.186 m The object looses energy in its way back also. If it makes it to the incline with excess kinetic energy, it will climb to a height less than its initial value. To calculate the final height the object reaches, we need to calculate the total energy lost to friction in both ways. Then using conservation of energy: W f = 2µ k mg (x + d) (10) E initial + W f = E final mgh 2µ k mg (x + d) = mgh final h final = h 2µ k (x + d) = 0.275 m!!! (11) But the height can not be negative. This means the object will loose all its energy and stop before it makes it back to the bottom of the incline! 2
Solution for Worksheet #06 3. A particle of mass m = 2.0 kg moves in a region of space with the following potential energy function. 60 P E(J) 40 20 0. 20 40 E for part e E for parts b, c and d 60 0 2 4 6 x(m) 8 10 12 (a) Find the work done on the particle when the particle moves from x = 4 m to x = 5 m. Answer: W = P E = [P E (4 m) P E (5 m)] = [20 J ( 10 J)] = 30 J (b) Assume the particle is initially at x = 11 m and has a velocity v 0 = 2 5 m/s in the positive x direction. Find the total mechanical energy of the particle. Answer: Initial potential energy (at x = 11 m) is P E = 0 J. Initial kinetic energy: KE = 1 2 mv2 = 1 2 (2.0 kg) ( 2 5 m/s ) 2 = 20 J Mechanical energy: E = KE + P E = 20 J. The mechanical energy in these conditions is the red horizontal line in the figure. (c) Where is the speed of the particle minimum and maximum? Answer: The speed and the kinetic energy are minimum (maximum) whenever the potential energy is maximum (minimum)...within the region allowed by the mechanical energy. This last part means that in our case the particle will move between x = 7 m and x = 12 m only. So the maximum speed happens at 10 m: KE = E P E (10 m) = 20 J ( 20 J) = 40 J but KE = 1 2K 2 (40 J) 2 mv2 v = m = 2.0 kg = 6.3 m/s And the minimum speed happens at the two turn-around points where the speed is zero, x = 7 m and x = 12 m. (d) Describe qualitatively the motion of the particle. Answer: The particle oscillates between x = 7 m and x = 12 m. The particle has zero speed at the turn around points and is subject to a force that pulls it toward the minimum 3
Solution for Worksheet #06 at x = 10 m. So the particle speeds up as it moves from the turn around point to the minimum and then slows down as it travels from the minimum to the other turn around point. (e) Now imagine that the total mechanical energy of the particle is E = 40 J. Describe the motion of the particle. Where is the speed of the particle maximum/minimum? Determine these speeds. Answer: Now the system oscillates between x = 1 m and x = 12.5 m, but the oscillations are more complex than before, because there are two Potential Energy minima. If the particle begins at x = 1 m, for instance, it will speed up until x = 3 m, slow down from there to x = 6 m, speed up again from x = 6 m to x = 10 m, and then slow down and come momentarily to a stop at x = 1 m. It then begins moving in the negative x direction in a similar way. The speed of the particle is maximum at the lowest minimum of the potential energy function, i.e. at x = 10 m. There, KE = E P E (10 m) = 40 J ( 20 J) = 60 J but KE = 1 2K 2 (60 J) 2 mv2 v = m = 2.0 kg = 7.7 m/s The minimum speed happens at the two turn-around points, x = 1 m and x = 12.5 m, where potential energy is equal to total energy and, hence, kinetic energy and the speed is zero. 4