Physics-Mechanics
A uniform bar is lying on a flat table. In addition to its weight and the normal force exerted by the table (which exactly balances the bar's weight), exactly two other forces, F1 and F2, act on the rod. If the net force acting on the rod is zero, then: A. the net torque will be zero if F1 and F2 are applied at the same point on the rod. B. the rod cannot accelerate translationally or rotationally. C. the rod can accelerate translationally if F1 and F2 are not applied at the same point on the rod. D. the net torque on the rod must also be zero.
A "The net torque on the rod must also be zero" is false: Fnet = 0 does not automatically imply that net = 0. Since the net torque need not be zero, there could be rotational acceleration, so "the rod cannot accelerate translationally or rotationally" can be eliminated. And "the rod can accelerate translationally if F1 and F2 are not applied at the same point on the rod" is false because we are told in the question that the net force on the rod is zero; this does automatically imply that the translational acceleration must be zero (regardless of where the forces are applied). The correct answer must be "the net torque will be zero if F1 and F2 are applied at the same point on the rod". (You might find it instructive to draw a quick sketch and try placing two forces F1 and F2, with the same magnitude and exactly opposite directions [because F1 + F2 = 0] at the same point on an object, and get a nonzero net torque. It can't be done.)
A pulley system is used in an auto repair shop to lift a 200 kg engine block 1.5 m out of a car. The lower pulley is attached directly to the engine block, and there are 8 strands of rope attaching the lower pulley to the upper pulley. The repairman pulls through 12 m of rope in the process of lifting the engine block. How much work is done during this process? Question 2 Answer Choices A. 3000 J B. 24 kJ C. 300 J D. 375 J
A One critical fact to remember about the mechanical advantage supplied by simple machines (like systems of pulleys) is that while they reduce the force necessary to lift a mass, they do not change the work done in the process. Therefore, W = mgh = 200 kg × 10 m/s2 × 1.5 m = 3000 J. In this case, the repairman must pull eight times the length of rope, applying one eighth the force, to lift the engine block as he would if he simply tied a rope directly to the block and lifted it straight up 1.5 m.
A housefly of mass m is sitting on the horizontal blade of a ceiling fan and has a tangential speed of v as the fan spins. The coefficient of static friction between the fly and the blade is μs and the acceleration due to gravity is g. As the fly walks toward the center of the fan, which of the following is true? Question 11 Answer Choices A. The force of static friction decreases and the fly stays on the fan. B. The force of static friction decreases and the fly slides off the fan. C. The force of static friction increases and the fly stays on the fan. D. The force of static friction increases and the fly slides off the fan.
A The centripetal force on the fly is the force of static friction so FCentripetal = FStatic Friction and mv2/r = FStatic Friction where r is the distance from the fly to the center of rotation of the fan. As the fly walks toward the center of the fan, r decreases. However, v also decreases since v is equal to r times the angular speed (which is constant at all points of a rotating object). The centripetal force therefore decreases. Since static friction has a maximum value but not a minimum value, it can decrease accordingly, and the fly will remain on the fan blade. (Note that mass in this case remains constant, and acceleration due to gravity is constant, eliminating those two answer choices).
At the beach, Kate collects sea shells of mass 100 g and places them in a bucket. She swings the bucket in a large circle (radius 0.4 m), vertical with the ground, so that the bucket swings over her head. When the bucket is at the top of the circle, upside down, the tangential velocity of the shells is 4 m/s. The sea shells do not fall out of the bucket. What is the net force on the sea shells inside the bucket at this point? Question 5 Answer Choices A. The net force is the same as the centripetal force. B. The net force is the same as the force of gravity. C. The net force is zero because the force of gravity cancels the centripetal force and the shells do not fall out of the bucket. D. The net force is the centripetal force plus the force of gravity since both are acting downward when the bucket is at the top of the circle.
A. The bucket is traveling in a circle, so the net force is the same as the centripetal force and is acting towards the center of the circle. By definition, the centripetal force is the net force acting towards the center of the circle. Both the force of gravity and the normal force from the bottom of the bucket are acting downward, creating the centripetal force, so gravity alone is not the centripetal force, eliminating two answer choices. The shells are traveling in a circle so their velocity is changing direction and they have non-zero acceleration, therefore the net force cannot be zero, eliminating the remaining incorrect answer choice.
A race car is traveling around a banked curve on a racetrack, meaning the road is not flat, but at an angle with the horizontal. Why can cars travel faster on a banked racetrack? Question 8 Answer Choices A. The normal force contributes to the centripetal force since the ground is at an B. The friction force is increased since the ground is at an angle, so the cars are less likely to slip. C. The force of gravity is decreased since the ground is at an angle, so the cars are less likely to slip. D. The normal force is increased since the ground is at an angle, so the cars are less likely to slip.
A. The centripetal force on the cars is the force that keeps the cars traveling in a circle and not flying off the track. On a flat road, the centripetal force is the force of static friction. On a banked road, the centripetal force is comprised of static friction and a component of the normal force. Since it is not friction alone that keeps the cars in a circle, they can travel faster without slipping. The banked road decreases the normal force, and the friction force, eliminating the answer choices stating friction or normal force increase. The force of gravity does not depend on the angle of the ground, eliminating the answer choice stating it decreases.
A housefly of mass m is sitting on the horizontal blade of a ceiling fan and has a tangential speed of v as the fan spins. The coefficient of static friction between the fly and the blade is μs and the acceleration due to gravity is g. As the fly walks away from the center of the fan, which of the following must be true so that the fly does not slip off the fan? Question 10 Answer Choices A. The force of static friction decreases. B. The force of static friction increases. C. The mass increases. D. The acceleration due to gravity increases.
B The centripetal force on the fly is the force of static friction so FCentripetal = FStatic Friction and mv2/r = FStatic Friction where r is the distance from the fly to the center of rotation of the fan. As the fly walks away from the center of the fan, r increases. However, v also increases since v is equal to r times the angular speed (which is constant at all points of a rotating object). The centripetal force therefore increases and the force of static friction must increase. (Note that mass in this case remains constant, and acceleration due to gravity is constant, eliminating those two answer choices).
A housefly of mass m is sitting on the horizontal blade of a ceiling fan and has a tangential speed of v as the fan spins. The coefficient of static friction between the fly and the blade is μs and the acceleration due to gravity is g. What is the maximum distance the fly can be from the point of rotation of the fan? Question 7 Answer Choices A. μsg/v2 B. v2/ μsg C. v/ μsg D. v2/ μsmg
B. The centripetal force on the fly is the force of static friction so FCentripetal = FStatic Friction. Since the question asks for the maximum distance, the centripetal force and force of static friction should be maximized so mv2/r = μsmg and r = v2/μsg (note that v increases proportionately to r, so the v2 term means that centripetal force increases with distance from the center).
An object weighing 100 N is traveling horizontally with respect to the surface of the earth in the absence of air resistance at a constant velocity of 5 m/s. What is the power required to maintain this motion? Question 5 Answer Choices A. 20 W B. 200 W C. 0 W D. 500 W
C 0 W is required to maintain this motion since there is no force component parallel to the direction of displacement, therefore no work is being done and no power is being exerted.
A 2.5-kg mass is projected straight upward with an initial kinetic energy of 980 J. If air resistance is ignored, how much kinetic energy will this projectile have as it strikes the ground? Question 17 Answer Choices A. 1470 J B. 1960 J C. 980 J D. 490 J
C Since air resistance is to be ignored, the original 980 J of kinetic energy is converted into 980 J of potential energy at the top of its path, which is in turn converted back into 980 J of kinetic energy as it strikes the ground. (The value given for the mass of the object is irrelevant.)
A car is traveling around a banked curve on a road, meaning the road is not flat, but at an angle with the horizontal. The car hits a patch of ice (static friction force is zero) on the road and slides at constant velocity around the curve. Is it possible for the car to slide around the curve and stay on the road? Question 9 Answer Choices A. No, the car will slide off the road without any static friction to provide the centripetal force. B. Yes, the car will stay on the road if it is heavy enough to provide the centripetal force. C. Yes, the car will stay on the road if the bank is steep enough for the normal force to provide the centripetal force. D. No, the car will slide off the road if it is heavy enough to provide the centripetal force.
C. The centripetal force on the car is the force that keeps the cars traveling in a circle and not flying off the road. On a flat road, the centripetal force is the force of static friction. On a banked road, the centripetal force is typically comprised of static friction and a component of the normal force. In this case, the car is on ice, so the centripetal force is comprised of a component of the normal force only. If the bank is angled enough, the component of the normal force pointing toward the center of the curve can be enough to provide the centripetal force for the car to stay on the road. Using N as the normal force, θ as the angle of the bank with the horizontal, m and v as the mass and velocity of the car, and r as the radius of curvature, yields Nsinθ = mv2/r as the equation to determine the necessary angle for the bank in order for the car to stay on the road. The weight of the car does not contribute to the centripetal force, eliminating two answer choices.
An object weighing 100 N is traveling vertically upward from the earth in the absence of air resistance at a constant velocity of 5 m/s. What is the power required to keep the object in motion? Question 4 Answer Choices A. 20 W B. 0 W C. 200 W D. 500 W
D In order to maintain a constant velocity pointing upward, the net force in the vertical direction must be zero. Since the force of gravity is exerting a downward force of 100 N, a 100 N upward force must be provided to maintain no net vertical force. With this knowledge, we may use the equation P = Fv. This gives P = (100 N)(5 m/s) = 500 W.
A carousel ride starts at rest. It takes 1 minute to reach its full rotation speed, then spins at a constant speed for 3 minutes. Finally, the carousel takes 1 minute to come to a stop again. Which best describes the forces acting on a person riding the carousel? I. During the first minute, there is a non-zero net force causing the person on the carousel to speed up. II. During the middle three minutes, there is zero net force since the speed is constant. III. During the last minute, there is a non-zero net force causing the person on the carousel to slow down
I and III During the first minute, there is a change in speed, so there is a non-zero acceleration, so there is a non-zero net force. Item I is correct. During the last minute, there is a change in speed, so there is a non-zero acceleration, so there is a non-zero net force. Item III is correct. During the middle three minutes, the speed is constant, but the direction of the person on the carousel is constantly changing in order to continue traveling in a circle. Since a change in direction is a change in velocity, and a change in velocity is by definition a non-zero acceleration, then there is also a non-zero net force acting to keep the person traveling in a circle. Item II is false. The correct answer is I and III.
A hammer is used to drive a nail into a board. Work is done in the act of driving the nail. Compared to the moment before the hammer strikes the nail, the mechanical energy of the hammer after its impact will be: Question 6 Answer Choices A. less, because work has been done on the hammer. B. greater, because work has been done on the hammer. C. greater, because the hammer has done work. D. less, because the hammer has done work
The hammer has done positive work on the nail, giving it enough energy to move into the wood. Thus, some of the hammer's original kinetic energy is lost, leaving it with less mechanical energy.
As a crate (of mass m) slides down a frictionless incline (with angle θ), a constant horizontal force F, parallel to the base of the incline, is applied to the crate so that the crate's speed down the incline remains constant. F = mg tan θ If the vertical rise of the incline is h meters, determine the work done by the horizontal force F as the crate slides down the incline. Question 9 Answer Choices A. -mgh sin θ cos θ B. -mgh sin2 θ C. -mgh D. -mgh cos2 θ
c Alternatively, one can use the relationship between work and gravitational potential energy to solve this problem. Since the object is traveling at a constant speed in the same direction, the net force on the crate is zero, and thus the component of the force of gravity going down the plane is equal to the component of the applied force going up the incline. The force of gravity is doing positive work on the crate, but since the kinetic energy of the crate is not changing, the applied force is doing the same magnitude of work on the crate, but negative. Remember that the work done by gravity is equal to the opposite change in gravitational potential energy. The crate lost mgh joules of gravitational potential energy, meaning gravity did mgh joules of work. Therefore, the applied force did -mgh joules of work.
