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Power AP  Physics 1 FRQ

Power AP  Physics 1 FRQ – Exam Style Questions etc.

Power AP  Physics 1 FRQ

Unit: 3. Work , Energy and Power

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Exam Style Practice Questions, Power AP  Physics 1 FRQ

Question

A ski lift carries skiers along a 600 meter slope inclined at 30°. To lift a single rider, it is necessary to move 70 kg of mass to the top of the lift. Under maximum load conditions, six riders per minute arrive at the top. If 60 percent of the energy supplied by the motor goes to overcoming friction, what average power must the motor supply?

Answer/Explanation

Ans:

6 riders per minute is equivalent to 6x(70kg)*9.8 = 4116 N of lifting force in 60 seconds
Work to lift riders = work to overcome gravity over the vertical displacement (600 sin 30)
Work lift = Fd = 4116N (300m) = 1.23×106 J
P lift = W / t = 1.23×106 J / 60 sec = 20580 W
But this is only 40% of the necessary power.           →        0.40 (total power) = 20580 W
                                                                                                         Total power needed = 51450 W

Question

The Sojourner rover vehicle shown in the sketch above was used to explore the surface of Mars as part of the Pathfinder mission in 1997. Use the data in the tables below to answer the questions that follow.
Mars Data                                                                                                         Sojourner Data
Radius: 0.53 x Earth’s radius                                                                        Mass of Sojourner vehicle:                                 11.5 kg
Mass: 0.11 x Earth’s mass                                                                              Wheel diameter: 0.13 m
                                                                                                                            Stored energy available:                                       5.4 x 105 J
                                                                                                                            Power required for driving
                                                                                                                            under average conditions:                                     10 W
                                                                                                                             Land speed:                                                             6.7 x 10-3 m/s
a. Determine the acceleration due to gravity at the surface of Mars in terms of g, the acceleration due to gravity at the surface of Earth.
b. Calculate Sojourner’s weight on the surface of Mars.
c. Assume that when leaving the Pathfinder spacecraft Sojourner rolls down a ramp inclined at 20° to the horizontal. The ramp must be lightweight but strong enough to support Sojourner. Calculate the minimum normal force that must be supplied by the ramp.
d. What is the net force on Sojourner as it travels across the Martian surface at constant velocity? Justify your answer.
e. Determine the maximum distance that Sojourner can travel on a horizontal Martian surface using its stored energy.
f. Suppose that 0.010% of the power for driving is expended against atmospheric drag as Sojourner travels on the Martian surface. Calculate the magnitude of the drag force.

Answer/Explanation

Ans:

(a) Plug into g = GMplanet / rplanet2 lookup earth mass and radius
gmars = 3.822 m/s2              to get it in terms of gearth divide by 9.8                        gmars = 0.39 gearth
(b) Since on the surface, simply plug into Fg = mg = (11.5)(3.8) = 44 N
(c) On the incline, Fn = mg cos θ = (44) cos (20) = 41 N
(d) moving at constant velocity → Fnet = 0
(e) W = P t           (5.4×105 J) = (10 W) t                      t = 54000 sec
d = v t                    (6.7×10-3)(54000 s)                        d = 362 m
(f) P = Fv               (10) = F (6.7×10-3)                         Fpush =1492.54 N total pushing force used
                                                                                           * (.0001) use for drag
                                                                                            → Fdrag = 0.15 N

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