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1 Jumping problems: Two phases: Jump with forces from gravity and leg muscles, acceleration upward. Flight, animal has left ground, gravity only force, acceleration downward. Treat these separately, each with it s own acceleration. Final velocity from jump is the initial velocity for the flight. P.53. When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.50 mm. a) What is the flea s acceleration during the jump phase? b) How long does the acceleration phase last? c) If the flea jumps straight up, how high will it go? Ignore air resistance here, which in reality would be very large and the flea will not reach this height. Prepare: The motion has two parts: The jump, with an acceleration upward, and the flight after the flea leaves the ground, with a downward acceleration. We treat these parts separately and use a constant acceleration during the jump (which would be the average acceleration in reality). Fleas are amazing jumpers; they can jump several times their body height something we cannot do. We assume constant acceleration so we can use the equations in Table.4. The last of the three relates the three variables we are concerned with in part (a): speed, distance (which we know), and acceleration (which we want). = a y y In part (b) we use the first equation in Table.4 because it relates the initial and final velocities and the acceleration (which we know) with the time interval (which we want). = a y t Part (cs about the phase of the jump after the flea reaches takeoff speed and leaves the ground. So now it is, that is 1.0 m/s instead of. And the acceleration is not the same as in part (a) it is now g (with the positive direction up) since we are ignoring air resistance. We do not know the time it takes the flea to reach maximum height, so we employ the last equation in Table.4 again because we know everything in that equation except y. Solve: (a) Use = 0.0 m/s and rearrange the last equation in Table.4. a y y (1.0 m/s) (0.50 mm) 1000 mm 1 m 1000 m/s (b) Having learned the acceleration from part (a) we can now rearrange the first equation in Table.4 to find the time it takes to reach takeoff speed. Again use =0.0 m/s. t a y 1.0 m/s 1000 m/s.0010 s (c) This time =0.0m/s as the flea reaches the top of its trajectory. Rearrange the last equation in Table.4 to get y v y i a y (1.0 m/s) ( 9.8 m/s ) m 5.1 cm Assess: Just over 5 cm is pretty good considering the size of a flea. It is about 10 0 times the size of a typical flea. Check carefully to see that each answer ends up in the appropriate units. The height of the flea at the top will round to 5. cm above the ground if you include the cm during the initial acceleration phase before the feet actually leave the ground.
2 dimensional problems Leafy spurge disperses its seeds with an explosive ejection from the seed capsule and can expel its seeds to distances up to 5 meters. The seeds have a mass of 3.04 mg each and are accelerated through a distance of mm. The seed pods are at the top of the plants which are up to 1 meter tall. Assuming the optimal 45º angle for the maximum distance, find the velocity with which the seed is ejected to reach the 5 meter range. Ignore air resistance (which is likely a poor assumption). Knowns: y o = 1m x o = 0 θ = 45º y f = 0 x f = 0 a = g in flight Because of the change in height, we cannot use the range formula from lab and the maximum height is no longer reached at half the time for the entire flight; the symmetry of the trajectory is destroyed. Instead, we will split the motion into vertical and horizontal components and solve them independently. Vertical: a = g, v yo = v o sin(45), y o = 1m, y f = 0 y = y o + v yo t + ½ (g) t = 1m + v o sin(45) t g/ t Horizontal: a=0, v xo = v o cos(45), x o = 0, x f = 5m x = x o + v xo t = v o cos(45) t v x = v o cos(45) v y = v yo + a y t = v o sin(45)  gt When the seed hits the ground at t f, y = 0 = 1m + v o sin(45) t f g/ t f x = 5m = v o cos(45) t f solve for t f (simpler than quadratic) drop units to save space, equations consistent. Use t f from x equation, drop units: 0 = 1 +( v o / )(5 /v o ) (g/)(5 /v o ) = g/v o t f = 5m/ v o cos(45) = 5 / v o sin(45)=cos(45)=1/ 5g/v o = 6 v o = 5g/6 = 5(9.8)/6 v o = 6.4 m/s t = 5 / v o = (5 [m]) / (6.4 [m/s]) = 1.1 s We can now determine the acceleration of the seed as it is ejected from the pod.
3 During ejection, the velocity changes from 0 to 6.4 m/s over a distance of mm. If we initially ignore gravity, we can use v f = v o +ad a = (v f  v o )/d = 6.4 /[(.00)] = 1.0 x 10 4 m/s Comparing this acceleration to g, it appears that ignoring g was justified, since it is much smaller than the value calculated. Using Newton s second law, we can calculate the force applied to the seed to eject it. F = ma = (3.04 x 106 kg)(1.0 x 10 4 m/s ) = 3.10 x 10  kg m/s = 3.10 x 10  N Ramp problems: Consider a block on a ramp making an angle of 30º with the horizontal. Under normal conditions, the block will move along the ramp surface. The coordinates which make the calculations the simplest are parallel to the ramp and perpendicular to it. This follows the general strategy of Choose one coordinate axis along the direction of the acceleration or motion and the other perpendicular to that. This means that the gravitational force needs to be split into components. I have given you some geometry tools on the second exam data sheet, anticipating that the geometry you learned in high school probably did not all stay in your ready recall memory. Applying these tools to the gravitational force (mg) gives the components in the diagram to the right. The gravitational force can be replaced by tits components, redrawing the forces with the ramp direction horizontal gives the diagram on the next page.
4 Unless the block is sinking into the ramp or jumping up off it, the acceleration perpendicular to the ramp is zero, as is the velocity. The acceleration along the ramp is downward or to the right in the diagram. Force balance: Perpendicular F N  mg cos(30) = 0 or F N = mg cos(30) Along ramp: F net = mg sin(30) f =ma (a) If the ramp is frictionless, then Fnet = mg sin(30) and the acceleration of the block is given by a = g sin(30) = g/. This gives v = v o + a t = v o + (g/)t and d = d o + v o t + ½ a t = d o + v o t + ½ (g/) t For displacement, take d o = 0 = v o t + ½ (g/) t (b) If there is friction, the friction force can be represented by f = μ F N where μ is the coefficient of friction. Let s see what we can determine from this and our force balance equations above. f = μ F N and, from the perpendicular force balance, F N = mg cos(30) giving f = μ mg cos(30) Then F net = mg sin(30) f = mg sin(30) μ mg cos(30) = mg [sin(30) μ cos(30)] = ma or a = g [sin(30) μ cos(30)]. Does this suggest a way to measure the coefficient of static friction (μ s ) when the block is standing still on the ramp?
5 1. A student brings in her physics homework, dragging it across campus in a wooden box. She is pulling with a force of 450 N on a rope, which inclined at 38 degrees to the horizontal. The ground exerts a horizontal force of 15N that opposes the motion. Calculate the acceleration of the crate if (at mass is 310 kg and (bts weight is 310 N. First, the crate will move (be acceleratedn the horizontal direction. Use coordinate axes which are horizontal and vertical. Resolve the 450N pull into horizontal and vertical components. The simplified force diagram is now: Vertical: F N + 450sin(38) = mg Horizontal: 450cos(38) 15 = ma a) m = 350 kg: a (450cos(38) 15) N 350kg 0.66 m / s b) wt = 350N = mg m = 350/g = 35.7 kg a (450cos(38) 15) N 35.7kg 6.43 m / s You can also calculate the Normal force from the ground: a) F N + 450sin(38) = mg = (350 kg)(9.8m/s ) = 630 N F N = sin(38) = 353 N b) F N + 450sin(38) = mg = 350 N F N = sin(38) = 73 N
6 . m1 is 3.70 kg on a frictionless inclined plane. Pulley is massless, and frictionless. m is.30 kg. Show force diagrams for each mass. Key ideas for solving text problems: (a) What is magnitude of acceleration of each block? (b) What is direction of acceleration of hanging block, m? (c) What is value of the tension in the cord? (1) The tension in a single piece of rope is the same anywhere in the rope and at both ends, as long as it is massless or nearly so. () The only effect of a massless, frictionless pulley is to change the direction the force, it does not affect the size of the force. of (3) Objects connected by string will have the same acceleration. (What would happen if this were not true?) From the diagram of the block on the ramp, the simplest axes are going to be along the ramp and perpendicular to it (you only need to break up one vector that way). This gives the following diagram with the components of the weight (force of gravityncluded: For this block, and F N = m 1 g cos(θrom the vertical balance T m 1 g sin(θ) = m 1 a along the ramp. For the hanging block, m g T = m a For connected objects like this, it is frequently fruitful to look at the entire set of connected objects as one total mass and set up the forces for that. Here, that gives m g m 1 g sin(θ) = (m 1 + m )a Which will allow us to solve for the acceleration a. Let s look at the forces along the ramp and for the hanging mass. T m 1 g sin(θ) = m 1 a m g T = m a We want to solve for tension T and acceleration a, so let me rearrange these to give T  m 1 a = m 1 g sin(θ) T + m a = m g Subtracting the first equation from the second leaves the equation above for the two masses together: m a + m 1 a = m g m 1 g sin(θ) (The T s subtract out). Once we solve this for a, we can use that value in either equation above to find the tension T.
7 If it is helpful, there is a not on the web site describing various ways to work with two equations in two unknowns. Some miscelaneous force diagrams:
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