Texas Tech University, Department of Physics
Physics 1308 Section 004


Part 1: Potential Energy and the Work Energy Theorem

Problem 1: Potential energy and a roller coaster. A 1000-kg roller coaster car moves from point A to B then C.
(a) What is the potential energy at B and C relative to A?
Solution:
                    The reference point for this part is A, therefore our y = 0 is at point A
                    Point B is a distance yb = -(30 - 10)m = -20 m below our reference point
                    Ub = mgyb = (1000 kg)(9.80 m/s2)(-20 m) = -196 kJ
                    Point C is a distance yc = -(30 - 25)m = -5 m below our reference point
                    Uc = mgyc = (1000 kg)(9.80 m/s2)(-5 m) = -49 kJ

(b) What is the change in potential energy as it goes from B to C?
Solution:
                    DU = Uc- Ub = -49 kJ - (-196 kJ) = 147 kJ

(c) What is the potential energy at B and C relative to point C?
Solution:
                    The reference point for this part is C, therefore our y = 0 is at point V
                    Point B is a distance yb = -(25 - 10)m = -15 m below our reference point
                    Ub = mgyb = (1000 kg)(9.80 m/s2)(-15 m) = -147 kJ
                    Point C is a distance yc = 0
                    Uc = mgyc = (1000 kg)(9.80 m/s2)(0) = 0

(d) What is the change in potential energy as it goes from B to C?
Solution:
                    DU = Uc- Ub = 0 kJ - (-147 kJ) = 147 kJ
                    Therefore, regardless of our choice of reference point, the change in potential energy between 2 points will always be the same.



Problem 2: A spring with spring constant k = 400 N/m, is stretched from it's equilibrium position by 1.5 m, find the energy stored in the spring. If the same spring is compressed by an amount of 1.5 m from the equilibrium, what would the energy be?
Solution:
                     Us = ½ kx2 =½(400 N/m)(1.5 m)2 = 450 J regardless of whether it was compressed by 1.5 m or stretched by 1.5 m.




 PART 2: Conservation of Energy and Conservative Forces

Problem 1: A dart of mass m = 0.100 kg is pressed against a spring of a toy dart gun. The spring has a spring constant k = 250 N/m, and it is compressed by a distance of xi = 0.06m. If the dart detaches from the spring once the spring reaches it's equilibrium position, xf=0, what is the speed of the dart?
Solution:
                    The initial energy of the system is purely elastic potential energy since the spring is compressed and the mass is initially at rest.  Therefore:
                            Ei = ½ kx2 =½(250 N/m)(0.06 m)2 = 0.45 J
                    The final energy of the system is purely kinetic since the mass is now moving and the spring is now back at its equilibrium position.  Therefore:
                            Ef = ½ mv2 =½(0.100 kg)v2
                    By conservation of energy:  Ei = Ef  therefore
                            ½(0.100 kg)v2 = 0.45 J
                            v2 = 2(0.45 J)/(0.100 kg)
                           v = 3.0 m/s



Problem 2: Assume the height of the roller coaster (see fig) is y = 40 m (take the reference point, y = 0, as the bottom of the coaster hill). If the car starts from rest at point A, what is the speed of the car at the bottom of the hill? What is the speed of the car when it reaches point B?
Solution:
                    The energy at A is purely gravitational potential energy since the car starts from rest.  Therefore:
                            EA = MgyA = M(9.80 m/s2)(40 m) = (392 M) J
                    The energy at the bottom of the hill is purely kinetic, no potential energy since y = 0.  Therefore
                            Ef = ½ Mv2 =½Mv2
                    Using conservation of energy, we can easily solve for v:
                            ½Mv2 = (392 M) J
                           v2 = 2(392 )
                           v = 28.0 m/s
                    The energy at B is a combination of potential and kinetic energy, therefore:
                            EB = ½ Mv2 + MgyB  = ½ Mv2 + (245 M)
                            EB = EA
                            ½ Mv2 + (245 M) = (392 M)   where     v2 = 2(392 -245)m2/s2
                            v2 = 294 m2/s2
                            v = 17.1 m/s



Problem 3: A ball of mass 2.00-kg is dropped from a height of 1.5 m (from the ground) onto a massless spring (the spring has an equilibrium length of 0.50 m). The ball compresses the spring by an amount of 0.20 m by the time it comes to a stop. Calculate the spring constant of the spring.
Solution:
                    First, we must pick a reference point.  The solid line across the diagram is our choice of y = 0.  The initial energy of the system is purely gravitational potential energy, and the final energy is a combination of gravitational potential energy ( it is 0.20 m below our reference point) and  elastic potential energy (the object stops momentarily as it fully compresses the spring).  The mass is originally at a height of 1.0 m above our reference point.  Therefore:
                            mg(1.0 m) = mg(-0.20 m) + ½ k(-0.20 m)2
                            (2.00 kg)(9.80 m/s2)(1.0 m) + (2.00 kg)(9.80 m/s2)(0.20 m) = 0.02 k
                            19.6 J + 3.92 J = 0.02 k
                            k = (23.5)/(0.02) = 1,180 N/m




 PART 3: Conservation of Energy and Non-conservative forces

Problem 1: A sled of mass m is given a kick on a frozen pond, imparting to it an initial speed vi = 2.0 m/s. The coefficient of kinetic friction between the ice and the sled is mk=0.10. Use energy considerations to find the distance the sled moves before stopping.
Solution:
                    The change in energy is equal to the work done by the non-conservative force:
                           DE = Ef - Ei = Wnc
                             Wnc= -fk d since the non-conservative force in this case is friction, therefore
                            d = (Ef - Ei)/(-fk)
                            fk = mkmg = (0.10)(mg)
                     The initial energy is purely kinetic and the final energy is 0:
                             Ef - Ei = 0 - ½mv2
                             d = (0 - ½mv2)/(-(0.10)(mg))
                                = (2.0)/(0.98)
                           d = 2.04 m



Problem 2: Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above the reference point. If the mass of the car is 1000-kg, and the car traveled a distance of 400 m, estimate the magnitude of the frictional force between the car and the track.
Solution:
                    Wnc =  -fk d
                    fk=(-DE)/d = (-Ef + Ei)/ 400 m = (-mgyb + mgya)/400 m
                       = (1000 kg)(9.80 m/s2)(40 m - 25 m)/(400 m)
                       = 370 N




 PART 4: Power

Problem 1: A 700 N marine in basic training climbs a 10.0 m vertical rope at a constant speed in a time of 8.0 s. What is his power output?
Solution:
                    In order for the marine to lift up the rope at a constant speed, he must exert a force upward equal to his weight.  The work he does is simple his weight multiplied by his displacement:
                            W = (700 N)(10.0 m) = 7000 J
                    The power output is just this work divided by the time it took him to reach the top:
                            P = W/t = (7000 J)/(8.0 s) = 875 W or about  880 W



Problem 2: A 200-kg crate is pulled along a level surface by an engine. The coefficient of friction between the crate and the surface is mk = 0.40.
(a) How much power must the engine deliver to move the crate at 5.0 m/s?
Solution:
                    The engine must work against the frictional force, therefore it has to exert a force along the direction of displacement equal to the frictional force.  Therefore:
                            F = mkmg = (0.40)(200 kg)(9.80 m/s2) = 784 N
                            P = F v = (784 N)(5.0 m/s) = 3,920 Watts

(b) How much work is done by the engine in 3 min. (3 min = 180 s)?
Solution:
                            P = W/t therefore, W = Pt
                            W = (3,920 J-s)/(180 s) = 706 kJ





ADDITIONAL PROBLEM: A 220 kg crate is lifted a distance of 21.0 m vertically with an acceleration of 1.47 m/s2 by a single cable.
(a) Find the tension on the cable
Solution:
                    The system is NOT in equilibrium since there is a net acceleration.  Therefore:
                            T - mg = ma
                            T = m (g + a) = (220 kg)( 9.80 + 1.47)
                           T = 2,480 N

(b) Find the net work done on the crate
Solution:
                    The net work done is done by the net force: Fnet = ma = (220 kg)(1.47 m/s2) = 323 N
                    Wnet = (323 N)(21.0 m) = 6,800 J

(c) the work done by gravity on the crate
Solution:
                    Wg = (Fg cos 180)(21.0 m) = -45,300 J

(d) Find the work done by the cable on the crate
Solution:
                   WT = T d = (2480 N)(21.0 N) = 52,100 J

(e) find the final speed of the crate assuming it started from rest.
Solution:
                    The werk-energy theorem states that the change in the kinetic energy of the system is equal to the net wok done.  The system starts from rest, therefore the final kinetic energy = 0.
                           DK = 1/2(220 kg)vf2
                            1/2(220 kg)vf2= 6800 J
                            vf2= 2(6800 J)/(220 kg) = 61.8
                            vf= 7,86 m/s



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