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1. What type of motion is ? Detail as much as possible.
2. What is the total displacement for the sum of ?
3. A bicycle decreases speed from to rest in 5 seconds. What’s its acceleration?
4. Under what conditions could acceleration be a positive number while an object decelerates?
5. What’s the sum of two velocity vectors, one of has a magnitude of in the positive y-direction, and the other pointing the opposite direction with a magnitude of ?
1. Answer: It’s a velocity, as determinable from the units. The magnitude (speed) is in the negative x-direction.
2. Answer: pointing 53° down from the x-axis. We find the total distance from the Pythagorean Theorem and recognize a 3-4-5 triangle with 32 + 42 = 52, so the length or magnitude of the vector is 5 cm, while the direction we find from . We need both magnitude and direction together to describe a direction. A picture works, too.
3. Answer: While we can’t determine direction, we can tell the magnitude of acceleration is . The negative sign and units are essential: -0.6 is a meaningless number and lost without them.
4. Answer: If the motion occurs in the negative direction, then positive acceleration would be deceleration because it's pointing away from the motion.
5. Answer: in the negative y-direction, or .
1. How far does an object travel in 3 seconds of free fall, measured in feet?
2. A 4 meter long ramp inclined at 10° from the horizontal has a grooved track on it for a ball, just as Galileo used. How long does it take for the ball to roll down the ramp starting at rest from the top?
3. A golfer drives the ball some 300 horizontal yards in 6 seconds. At what time is the velocity smallest and what is it?
4. A golfer drives the ball with an initial velocity of at an angle of 43° above horizontal. What is the final velocity in the y-direction, just before it hits the ground?
5. A golfer drives the ball with an initial velocity of at an angle of 42° above horizontal. How high does it travel before turning around?
1. Answer: Finding the answer in SI units first is the way to go, because we know the acceleration of gravity in those units but not those of feet. Using with , we find y = 44.1 m. To convert to meters, we use 1 foot is 0.3048 meter, so that’s a fall of 145 feet in three measly seconds. Gravity gets us every time.
2. Answer: In this case, a = g sin θ in . Solving for t, we first arrive at . See how those units cancel out and second move to the numerator? Units are a physicist’s best friend. Lastly, we take the square root for t = 2.2 s.
3. Answer: At time 3 seconds, the ball is at the peak of its motion, with no velocity in the y-direction in the moment of between rising up and falling down. The x-velocity, on the other hand, is constant, and therefore isn’t less at any particular time than another. Since we’re given the time the ball travels a constant horizontal speed over a horizontal distance, we can calculate that speed using SI units.
There are 0.9144 m in 1 yd, so 300 yards is . Now we’re ready to calculate .
4. Answer: It’s really too bad we’re not asked for the final composite velocity, because it would be the same but with a -43° angle. That’s how parabolas work: at the same vertical distance of 0 m the velocity is the same numeric value, as verified by the equation vfy2 = viy2 + 2aΔy. When Δy = 0, then vf2 = vi2. That’s not true for the x-direction alone because Δx isn’t 0.
What we really need to know for our problem is the initial velocity in the y-direction alone, vyi,and we’ll have it from vyi = 56 sin 43, or . Since the ball flips direction when as it returns to the ground, the final velocity .
5. Answer: Let’s make a picture to visualize the problem and its components. If anyone asks, we like drawing.
We only care about the y-direction in this problem to find Δy, when the acceleration of gravity brings the final velocity to 0 before turning around. We find the x- and y-components of velocity to be 56 cos 43 and 56 sin 43 respectively, so . We don’t need vx to answer this question, though we would if we needed the horizontal range.
The equation that would help us here is vfy2 = viy2 + 2aΔy, with g as because it’s in the opposite direction from initial velocity.
1. What is the reaction force of a hand pulling on a string?
2. An object rests on a bookcase. It’s probably a book. Anyways, if the normal force acting on this book is 10 N, what’s the mass of the book?
3. A skateboarder skids to a stop in 2 seconds. His original velocity was . What’s the magnitude of the decelerating force if the skateboarder plus board are 65 kg?
4. Complete a free-body diagram for a tug-of-war rope in use, stationary.
5. What’s the sum of the forces for a combination of , , , and ?
1. Answer: The string pulling the hand. Newton’s Third Law flips the two actors around for the reaction.
2. Answer: We can draw a FBD to identify forces and how they relate to each other.
The diagram informs us that N = W because . Alternatively, we could write the sum of the forces as N – W = 0. They’re equivalent. Since weight is always mg, we next write N = mg and therefore we solve for m = 1.02 kg.
3. Answer: We need both and F = ma. Either by subbing the former into the latter, or by finding acceleration numerically before sticking it into the force equation we find that F = -97.5 N. It’s negative if we call the direction of motion positive, anyway. Since we were asked for only magnitude and not direction, the positive value will do.
Note that the tension on each end of the rope has to equal the other side, because the net force in the x-direction is 0. Then, since the net force in the y-direction must be 0 as well, the tensions need to add to W = mg collectively.
5. Answer: Yes, this is more of vector practice than force practice, but it’s good practice so we can’t complain. We must add the x- and y-direction forces separately, for a net of and . Using the Pythagorean Theorem, the magnitude is , or 13 N, in the direction of = 22.6° below the negative axis, as shown below.
1. What distance separates two masses of 1 × 1010 kg each if the force between them is 100 N?
2. What’s the orbital velocity for the configuration in #1?
3.The orbital period for planet Shmoop’s moon is 24 hours, at a distance 1.4 × 107 m away. What is the acceleration of the moon?
4. A battery-powered toy car travels 5 cm per second. When fastened to a 1 m string with the end tacked to the floor, what’s the centripetal force on the 250 g car?
5. Draw a free-body diagram for planet Shmoop during a moment in time of its orbit around the star Educatia. Label the force, acceleration, and velocity vectors. Also label any forces on Educatia.
1. Answer: Using with , we arrive at r = 8,167 m. Well, we do if we are careful with the scientific notation. No, this doesn’t model any real configuration in outer space.
2. Answer: We know that for orbital motion, and therefore that . It doesn’t matter which of the masses we choose, though normally we’d choose the smaller of the two as the orbiting body. In reality, the motion of the larger body may also change and we’d have to consider the center of mass, which we’re not going to do right now, because we haven’t learned how just yet.
What we are going to do is calculate . Then we take the square root to find . Definitely not a practical orbital speed, but a practical problem with which to practice.
3. Answer: Since , we had better find ourselves the speed from the orbital period. We can do that from logic alone to find : the distance is 2πr, while the time is 24 hours times 60 minutes per hour, times 60 seconds per minute. It’s 86,400 seconds. Therefore, the speed is .
We’re all set to find the acceleration now with : it’s .
4. Answer: From . Going around in circles makes us dizzy because of forces on our brains! It makes sense now.
Because the force of gravity on each body is an action-reaction pair, they’re the same size but in opposite directions. Acceleration of the larger body is negligible because it’s not in orbit, being more massive. The acceleration of Shmoop points the same direction as the force of gravity because force and acceleration are BFFs. Velocity doesn’t hang with them: it’s perpendicular, tangential to the orbit.
Please refer to the diagram above for Questions 1 – 5.
1. What is the sum of the forces for this system of weights if neither block is moving, with a = 0?
2. How would one calculate the friction in this diagram if a = 0?
3. If the masses of the objects are 100 g for m1 and 80 g for m2 while a = 0, then what are the values of T and N?
4. If the masses of the objects are 100 g for m1 and 80 g for m2 with a = 0, then what is the coefficient of static friction?
5. How do the force equations change if acceleration is positive? Write each of them.
1. Answer: The sum of the forces must be zero if there’s no acceleration. That means that T = m2g, T = Ff + m1gsin 30, and N = m1g cos 30.
2. Answer: The friction is static friction, present when forces are insufficient to get an object moving, and calculable by Ff = μsN. Since N = m1g cos 30, we may add that Ff = μsm1g cos 30.
3. Answer: We know that T = m2g, T = Ff + m1gsin 30, and N = m1g cos 30 from the sum of forces being 0. We only need the first and last of these to find the numerical values of T and N. Therefore, , while .
4. Answer: We just found T and N in #3 (or if not, we could find them how we did in #3), so we plug T = 0.78 N and N = 0.85 N into T = Ff + m1gsin 30, rewritten as T = μsN + m1gsin 30. This leads directly, after some algebra, to μs = 0.34. No units necessary.
5. Answer: Instead of T = m2g, T = Ff + m1gsin 30, and N = m1g cos 30 that we had before with no net force, we have m2a = T − m2g for Block 2, m1a = Ff + m1gsin 30 – T for Block 1, and N = m1g cos 30 is still true because there’s no acceleration in that direction. Note that the force of friction becomes kinetic and we were appropriately careful about vector directions while writing these equations.