B LEVEL PROBLEMS

1)      What is the minimum work needed to push a 1000-kg car 300 m up a 17.70 incline? (a) Ignore friction. (b) Assume the effective coefficient of friction is 0.25.

SOLUTION:

The minimum work is needed when there is no acceleration, that is, when the system is in equilibrium.

      (a)  From the force diagram, we write SF = 0:

                  y-component: SFy = 0 therefore FN – mg cos q = 0;

                  x-component: SFx = 0 therefore Fmin – mg sin q = 0.

            For a distance d along the incline, we have

                  Wmin  = Fmind cos 0° = mgd sin q (1) 

                           = (1000 kg)(9.80 m/s2)(300 m) sin 17.5° 

                           = 8.8 ´ 105 J.

      (b)  When there is friction, the minimum force required increases, now you are acting against a component of gravity plus friction.  Therefore,

 

                  x-component: SFx = 0;      Fmin – mg sin q – mkFN = 0,or

                  Fmin = mg sin q + mkmg cos q= 0,

                  For a distance d along the incline, we have

                  Wmin    = Fmind cos 0° = mgd (sin q + mk cos q)(1) 

                             = (1000 kg)(9.80 m/s2)(300 m)(sin 17.5° + 0.25 cos 17.5°) =1.6 ´ 106 J.

 

 

 

 

 

2)      At an accident scene on a level road, investigators measure a car’s skid mark to be 88 m long. It was a rainy day and the coefficient of friction was estimated to be 0.42.use the data to determine the speed of the car when the driver slammed on (and locked) the brakes.(Why does the car’s mass not matter?)

SOLUTION:

On a level road, the normal force is mg, so the kinetic friction force is mkmg . Because it is the (negative) work of the friction force that stops the car, we have

            W = Dk;

            – mkmg d = ½mv2 – ½mv02;

            – (0.42)m(9.80 m/s2)(88 m) = – ½mv02, which gives v0 =27 m/s (97 km/h or 60 mi/h).

Because every term contains the mass, it cancels.

3)      A 55-kg hiker starts at an elevation of 1600 m and climbs to the top of a 3100-m peak. (a) What is the hiker’s change in potential energy? (b) What is the minimum work required of the hiker? (b) Can the actual work done be more than this? Explain why.

SOLUTION:

      (a)  With the reference level at the ground, for the potential energy change we have

                  DU = mg Dy = (55 kg)(9.80 m/s2)(3100 m – 1600 m) = 8.1 ´ 105 J.

      (b)  The minimum work would be equal to the change in potential energy:

                  Wmin =DU = 8.1 ´105 J.

(c)    Yes, the actual work will be more than this. There will be additional work required for any kinetic energy change, and to overcome retarding forces, such as air resistance and ground.

4)      A projectile is fired at an upward angle of 45.00 from the top of a 265-m cliff with a speed of 185 m/s. What will be its speed when it strikes the ground below? (Use conservation of energy).

SOLUTION:

We choose the potential energy to be zero at the ground (y = 0). Energy is conserved, so we have

            Ei = Ef, where the initial energy is BOTH kinetic and potential, while the final energy is only kinetic.

            ki + Ui = kf + Uf ;

            ½mvi2 + mgyi = ½mvf2 + mgyf ;

            ½m(185 m/s)2 + m(9.80 m/s2)(265 m) = ½mvf2 + m(9.80 m/s2)(0), which gives vf = 199 m/s.

      Note that we have not found the direction of the velocity.

5)      A small mass slides without friction along a looped apparatus as shown. If the object remains on the track, even at the top of the circle (whose radius is r), from what minimum height h must it be released?

SOLUTION:

 

      With y = 0 at the bottom of the circle, we call the start point A, the bottom of the circle B, and the top of the circle C.

      At the top of the circle we have the forces mg and FN, both downward, that provide the centripetal acceleration:

            mg + FN = mvC2/r.

      The minimum value of FN is zero, so the minimum speed at C is found from 

            vCmin2 = gr.

      From energy conservation for the motion from A to C we have 

            kA + UA = kC + UC ;   

            0 + mgh = ½mvC2 + mg(2r), 

      thus the minimum height is found from

            gh = ½vCmin2 + 2gr = ½gr + 2gr, which gives h = 2.5r.

6)      A 90-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15-m the floor is frictionless, and for the next 15-m the coefficient of friction is 0.30.What is the final speed of the crate?

SOLUTION:

On the level the normal force is FN= mg, so the friction force is Ffr = µkmg.

      For the work-energy principle, we have

            WNC = Ef – Ei = Dk + DU = (½mvf2 – ½mvi2) + mg(hf – hi);

            F(L1 + L2) –µkmg L2 = (½mvf2 – 0) + mg(0 – 0);

            (350 N)(15 m + 15 m) – (0.30)(90 kg)(9.80 m/s2)(15 m) = ½ (90 kg)vf2,

      which gives vf =12 m/s.

7)      During a workout, the football players at State U ran up the stadium stairs in 61 s. The stairs are 140 m long and inclined at an angle of 300.If a typical player has a mass of 105 kg, estimate the average power output on the way up. Ignore any frictional forces.

SOLUTION:

The work done increases the potential energy of the player. We find the power from

            P  = W/t = DU/t = mg(hf – hi)/t

                 = (105 kg)(9.80 m/s2)[(140 m) sin 30° – 0]/(61 s) = 1.2 ´ 103 W (about 1.6 hp).