Physics 6A Practice Midterm #2 solutions 1. A locomotive engine of mass M i

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Physics 6A Practice Midterm #2 solutions 1. A locomotive engine of mass M i

Carlos Albizu University, US has reference to this Academic Journal, Physics 6A Practice Midterm #2 solutions 1. A locomotive engine of mass M is attached so that 5 train cars, each of mass M. The engine produces a constant force that moves the train forward at acceleration a. If 3 of the cars are removed, what will be the acceleration of the shorter train? We just use Newton?s 2nd law here: Net force = (total mass) x (acceleration) Initially, the total mass is 6M (5 cars, plus the engine) After removing 3 cars the total mass is 3M (2 cars, plus the engine) In both cases the Net force is the same (the engine didn?t change). Here is the formula in both cases: Initial F = (6M)(a) Final F = (3M)(afinal) Setting these equal gives afinal = 2a Another way so that think about this one is that since the mass is cut in half, the acceleration must double. We can think of this as a single box alongside total mass 4kg. Then using F = ma we get the acceleration of the whole system. Now we do the same thing, but just in consideration of box B: Remember ? Box B has a mass of only 3kg 2. Two boxes are placed next so that each other on a smooth flat surface. Box A has mass 1 kg in addition to Box B has mass 3 kg. A constant horizontal force of 8 N is applied so that Box A. Find the force exerted on Box B.

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We can think of this as a single box alongside total mass 4kg. Then using F = ma we get the acceleration of the whole system. We need so that account in consideration of friction, so first find the maximum force of static friction: 8N Fstatic This friction force is more than the 8N force trying so that move the boxes, so they are held in place by friction. Thus the acceleration of both boxes is 0. *Note that the actual force of friction holding the boxes in place is only 8N ? just enough so that keep them from moving. 3. Two boxes are placed next so that each other on a flat surface. Box A has mass 1kg in addition to Box B has mass 3kg. The coefficients of friction are 0.3 in addition to 0.4 in consideration of kinetic in addition to static friction, respectively. A constant horizontal force of 8N is applied. Find the acceleration of Box B. 4. A crate filled alongside books rests on a horizontal floor. The total weight of the crate in addition to books is 700N. The coefficients of friction are 0.35 in consideration of kinetic in addition to 0.45 in consideration of static. A force of 450N is applied so that the crate at an angle of 20ø below the horizontal. Find the acceleration of the crate. We need so that find the components of the 450N force: We need so that find the normal force, so we can determine whether static friction will hold the crate in place. For this, we use the fact that the forces in the y-direction cancel out: This gives a maximum force of static friction = (0.45)(854N) = 384N Comparing this alongside the x-component of the applied force, we see that this is not enough friction so that hold the crate in place, so the actual friction force will be kinetic. Ffriction,k = (0.35)(854N) = 299N Now we can use F=ma again in consideration of the x-direction forces: 5. Blocks A in addition to B in addition to C are connected by a massless string in addition to placed as shown, alongside Block A on the upward slope of the 60ø incline, Block B on the horizontal surface at the top in addition to Block C on the downward slope of the 60ø incline. The string passes over a frictionless, massless pulley. The inclines are frictionless, but the horizontal surface has coefficient of kinetic friction 0.45. Blocks A in addition to B have mass 5kg. The system accelerates at 2 m/s2. Find the mass of Block C. Here is the force diagram in consideration of this problem. The blocks will move together, so we can consider the motion of the entire system in one formula. Our axis system will be chosen so that coincide alongside the string. With this choice, our formula is: Notice that the tension forces cancel out. We are assuming the string in addition to pulley are massless, so the tension is the same throughout the string. Thus the forward tension T1 on A is canceled by the backward tension T1 on B, in addition to similarly the T2 cancels out. Or we say that the tensions are internal forces, in addition to thus do not affect the acceleration of the entire system. mAg mCg mAgsin? NB

Now we need so that find the friction force. For this we need the Normal force on B. From the diagram we see that NB = mBg. Since the blocks are in motion, we are dealing alongside kinetic friction. Now we can substitute into the force formula: 5. Blocks A in addition to B in addition to C are connected by a massless string in addition to placed as shown, alongside Block A on the upward slope of the 60ø incline, Block B on the horizontal surface at the top in addition to Block C on the downward slope of the 60ø incline. The string passes over a frictionless, massless pulley. The inclines are frictionless, but the horizontal surface has coefficient of kinetic friction 0.45. Blocks A in addition to B have mass 5kg. The system accelerates at 2 m/s2. Find the mass of Block C. A bit of algebra will get us the desired mass mC 6. A 0.2 kg piece of wood is being held in place against a vertical wall by a horizontal force of 5 N. Find the magnitude of the friction force acting on the wood. 5N N mg Ffriction From the force diagram, we can see that the friction force must equal the weight of the piece of wood so that keep it from falling. 7. A 4 kg block on a horizontal surface is attached so that a spring alongside a force constant of 50 N/m. As the spring in addition to block are pulled forward at constant speed, the spring stretches by 25 cm. Find the coefficient of kinetic friction between the block in addition to the table. The key phrase here is ?constant speed?. Since the block is moving at constant speed (and direction) we know that its acceleration is 0. Thus the net force is 0 as well. Here is the force diagram. Since the net force is 0, we know that the Normal force must equal the weight, in addition to the Friction force must equal the Spring force. The spring force obeys Hooke?s Law: The weight is: Finally, we can put this together so that find the friction:

Physics 6B Standing Waves Standing Waves Standing Waves

8. A 1000 kg car is driven around a turn of radius 50 m. What is the maximum safe speed of the car if the coefficient of static friction between the tires in addition to the road is 0.75? Ffriction The friction force must be directed toward the center of the circle. Otherwise the car will slide off the road. If we want the maximum speed, then we want the maximum static friction force. The road is flat (not banked), so the Normal force on the car is just its weight. This friction force is the only force directed toward the center, so it must be the centripetal force: 9. A 90 kg man drives his car at a constant speed of 15 m/s over a small hill that has a circular cross section of radius 40 m. Find his apparent weight as he crests the top of the hill. (Hint: the apparent weight is the same as the normal force on the man.) 40 m 15 m/s When the car reaches the top of the hill, it will have only 2 forces: its weight, in addition to the normal force supplied by the road. Since the man is sitting in the car, he feels a normal force as well. The car (and man) must be accelerating toward the center of the circle, so the net force on the man will be equal so that the centripetal force required so that keep him moving along the circle. mg N 10. Planet X has a radius that is 3 times as large as Earth in addition to a mass that is 6 times that of Earth. A NASA astronaut who weighs 550 N here on Earth is planning so that embark on a manned mission so that Planet X. What will be the astronaut?s weight when she lands there? The weight is the same as the gravitational force. On Earth the weight is 550N: In this formula, G is constant, in addition to m will not change when the astronaut goes so that a new planet. So the only parts that will change are the mass in addition to radius of the planet. The easy way so that do this problem is so that simply substitute the new planet X values into the formula, in addition to rearrange it: Planet X Earth

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