Worksheet 6: Work and Energy

[Problem 1] A 1200-N crate rests on the floor. How much work is required to move it at a constant speed (a) 4.0-m along the floor against a friction force of 230-N?
Solution:
                    In order to move something against a constant frictional force, you must exert a force greater than or equal to the frictional force. So you exert at least F = 230 N in order to move the crate directed along the displacement direction. Therefore:
                    W = Fd cosq = (230 N)(4.0 m)(1) = 920 J (answer)

(b) 4.0-m vertically?
Solution:
                    This one is similar to the firefighter problem. The force you must exert must be at least equal to the weight of the crate.
                    W = 1200 N (4.0 m) = 4800 J (answer)



[Problem 2] 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 3] 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.


[Problem 4] 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 5] 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 6] 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 7] 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


[Problem 8] How long will it take a 1750-W motor to lift a 285-kg piano to a sixth story building 16.0-m high?
Solution:
                    First, find the work needed to lift the piano: W = mgh = (285 kg)(9.80 m/s2 )(16.0 m) = 44,688 J
                    P = W/t
                    t = W/P = (44,688 J)/(1750 W) = 25.5 s (answer).


[Problem 9] An 80.0-kg football player traveling 5.0 m/s is stopped in 1.0-s by a tackler. (a) What is the original kinetic energy of the player? (b) What average power is required to stop him?
Solution:
                    K = ½ mv2 = ½ (80.0 kg)(5.0 m/s)2 = 1000 J  (answer)
                    P = W/t = 1000 J/(1.0 s) = 1000 W  (answer)