for Solution Long-answer Homework 18 Motion in a Plane problems
Solution to Long-answer Homework Problem 18.1()
Problem: A box leaves the edge of a table (1.00m above the ground) with an initial velocity of 1.00 m in the s horizontal direction. (a)How long does it take to hit ground? (b)How far does the box travel horizontally before it lands? (c)What is the velocity of the box when it lands? (d)Suppose the box is put on an incline (with angle of inclination ) on the table top so that it can gain the velocity it needs to leave the table. As it leaves the table, there is a frictionless curvey bit that changes the direction of the velocity to horizontal, without changing its size. What is the vertical height up the incline at which the box must be released in order to have the desired initial horizontal velocity of 1 m when it leaves the table? You may assume that the incline is frictionless. s Solution to Part (a) The velocity of the box in the vertical direction is zero. To find the time the box takes to hit the ground, consider only the vertical direction, taking positive upward. 1 y = - gt2 2 t= 2(y) = -g 2(-1m) m = 0.451s -9.81 s2
Grading Key: Part (a) 5 Points Solution to Part (b) Now, only consider motion in the horizontal direction. The horizontal velocity doesn't change over the time the box falls. m (0.451s) = 0.451m x = vx t = 1.0 s Grading Key: Part (b) 3 Points Solution to Part (c) Now, we must consider the magnitude and direction of the velocity using the x- and y-components. vy = -gt = -9.81 m m (0.451s) = -4.43 s2 s m m x - 4.43 y ^ ^ s s
v = vx x + vy y = 1 ^ ^ Grading Key: Part (c) 4 Points
Solution to Part (d) 1
There are a number of ways to solve this problem; here's one. Acceleration down an incline is given by a = g sin (you can easily derive this by drawing a picture and decomposing the force of gravitation into components along the incline and perpendicular to the incline) This is the acceleration that the box experiences down the incline. Since we only know what the final velocity needs to be (to derive this you can eliminate t between the constant-acceleration equations for position and velocity) or, we can just use conservation of energy: The vertical distance needs to be such that -Ug = 1 mv 2 : 2 y = 1 v2 9.8 2 f = 5.1cm
Fn a
Fg
Fg
Something to think about: originally when I gave this problem as part of a test, I had put friction on this incline, right up to the curvey bit, so you had to figure out a loss of energy due to the normal force times the friction coeffficient times the distance down the incline related to the distance fallen. I gave both the angle and the coefficient of friction. Somehow this got lost in typing up the solutions. My guess is Jennifer got tired and didn't want to have to draw the free body diagram which was a good part of the points. Grading Key: Part (d) 4 Points Total Points for Problem: 16 Points
Solution to Long-answer Homework Problem 18.2()
Problem: A projectile is fired from the top of a building at an angle of 30 to the horizontal, with an initial speed of 50 m/s. The building is 75 m tall. (a)Break the initial velocity down into x and y components. (b)How long does it take for the projectile to reach the ground? (c)What is the velocity of the projectile the instant before it strikes the ground? (d)How far from the building (horizontally) does the projectile hit the ground? (e)How much work did the Earth do on the projectile, if its inertia is m = 1kg, if you consider just the projectile as your system? if you consider the Earth-projectile as your system? Solution to Part (a) v0x = v0 cos = v0y 3 v0 = 43.3 m/s 2 1 = v0 sin = v0 = 25.0 m/s 2 2
Grading Key: Part (a) 5 Points 2 point(s) : x and y components chosen as horizontal and vertical. 1 point(s) : x is cosine 1 point(s) : does math right 1 point(s) : defines directions taken as positive, at least for the y direction. If doesn't here, but does on next part, let them have this point. Solution to Part (b) The projectile starts at a height h with vertical velocity v0y , and is accelerated by gravity alone. So 1 y = v0y t - gt2 2 1 -h = v0y t - gt2 2 1 g t2 - (v0y )t - h = 0 2 Use the quadratic formula, t = Choose positive time, t = 1 v0y g
2 v0y + 2gh
1 v0y + g
2 v0y + 2gh
= 7.22 s
Grading Key: Part (b) 6 Points 1 point(s) : v0,y opposite sign of acceleration due to gravity 1 point(s) : a = g 2 point(s) : correct constant acceleration equation 1 point(s) : Correct answer with units Solution to Part (c) Vertical velocity at time t is given by vy (t) = v0y - gt = 25.0 m/s - (9.8 m/s2 )(7.22 s) = -45.8 m/s Horizontal velocity is unaccelerated, therefore vx (t) = v0x Grading Key: Part (c) 5 Points 1 point(s) : Reports as vector or components 2 point(s) : vf,x = v0,x NO ACCELERATION 1 point(s) : calculates vf,y from a correct relationship for uniform acceleration 1 point(s) : calculates it correctly 3 1 point(s) : v0 = |vy,0 | from part a
Solution to Part (d) The horizontal position at time t is given by (no horizontal acceleration) x = v0x t = (43.3 m/s)(7.22 s) = 313 m Grading Key: Part (d) 4 Points 2 point(s) : Correct zero acceleration equation for distance 1 point(s) : vx,0 from (a) 1 point(s) : t from (b) Solution to Part (e) Considering just the projectile as your system, using h = height of building, the work is mgh = 1kg 9.8 m 75m = 735J s2
Considering the Earth-projectile system, the work is zero. Grading Key: Part (e) 7 Points 4 point(s) : projectile only: 2 pts. for mgh. 2pts. for correct height, realizing only how far it changes matters. 3 point(s) : Earth-projectile system. IBC if the parts are backwards. Total Points for Problem: 27 Points
Solution to Long-answer Homework Problem 18.3()
Problem: A 5 kg block initially has a speed of 1.5 m/s at the top of a 30 incline, and slides a distance of 5.0 m down the incline in 1.75 s. (a)Find the magnitude of the acceleration of the block. (b)Draw a free-body diagram for the block. You may not neglect friction. Use a coordinate system with the x-axis pointed down the incline, and decompose all vectors into x and y components. Label everything. (c)Find the components of the gravitational force in the given coordinate system. (d)Find the normal force and the force of kinetic friction. (e)Find the coefficient of kinetic friction. (f)What is the final speed of the block? Solution to Part (a) Assuming constant acceleration, 1 x = v0 t + at2 2 2 m 2 (5m - (1.5 = m/s)1.75s) 1.55 2 a = 2 (x - v0 t) = t (1.75s)2 s 4
Grading Key: Part (a) 3 Points 1 point(s) : Constant acceleration equation. 1 point(s) : correct substitutions 1 point(s) : Correct answer. Solution to Part (b)
y x Ff FN = mg cos
F x = mg sin F y = mg cos F
g e,b g
g
a = mg
Grading Key: Part (b) 4 Points 1 point(s) : x-axis in direction of motion. 1 point(s) : reasonable relative magnitudes 1 point(s) : Correctly placed and labelled F N . 1 point(s) : Correctly placed and labelled F g . Solution to Part (c)
g Fx = mg sin = (5k)(9.8 m/s2 )
g Fy = -mg cos = 1(5k)(9.8 m/s2 )
1 = 24.5N 2 3 = -42.4N 2
Grading Key: Part (c) 2 Points 1 point(s) (2 times) : Each component, correct expression and/or number. 5
Solution to Part (d)
g There is no acceleration in the y-direction, so sum of forces is zero, so Fy = -F N .
F N = mg cos = 42.4N Se can find the force of kinetic friction using Newton's second law for the x-components,
g ma = Fx - F f g F f = Fx - ma = m(g sin - a) = (5kg)(g/2 - 1.55 m/s2 ) = 16.75N
Grading Key: Part (d) 6 Points 1 point(s) : Balance normal forces. 1 point(s) : F N , correct expression and/or number. 2 point(s) : Use Newton II for tangential forces. 2 point(s) : F f , correct final expression and/or number. Solution to Part (e) F f = k F N k = Ff m(g sin - a) 16.75N = = = 0.4 FN mg cos 42.4N
Grading Key: Part (e) 2 Points 1 point(s) : Correct expression. 1 point(s) : Correct answer. Solution to Part (f) vf = vi + at = 1.5 m/s + (1.55 m/s2 )(1.75s) = 4.2 m/s Grading Key: Part (f) 2 Points 1 point(s) : Expression (implied by correct answer). 1 point(s) : Correct answer. Total Points for Problem: 19 Points
Solution to Long-answer Homework Problem 18.4(Where's the Work?)
Problem: You have a 6kg cart that has a spring attached to the front. You hold the cart so the spring is compressed against the wall. You let go of the cart, and the spring pushes it away from the wall. The cart's center of mass moves 50cm before it loses contact with the wall, at which point it is traveling 3m/s. (a)What is the average force on the cart? (b)What is the total work done by the wall on the cart? 6
(c)What is the change in kinetic energy of the center of mass of the cart? Solution to Part (a) F = ma, so we need to find a. There are plenty of ways to do this, probably the easiest is to use
2 2 vf = vi + 2ax
(if you don't remember this, you can derive it by solving the velocity equation for t and plugging that into the position equation and simplifying) Solving for a with vi = 0, a=
2 vf 3m m s = =9 2 2x 2(0.5m) s 2
(in the direction of motion) So the average force is F = ma = (6kg) 9 Grading Key: Part (a) 2 Points Solution to Part (b) No work is done by the wall. This is a little subtle, but the wall does no work on the cart, since the point of application of the force does not move. The spring is basically doing work on the cart. Grading Key: Part (b) 2 Points Solution to Part (c) Calculating K directly, K = Kf - Ki = 1 m 1 2 mvf = (6kg) 3 2 2 s
2
m = 54N s2
= 27J
Grading Key: Part (c) 2 Points Total Points for Problem: 6 Points
Solution to Long-answer Homework Problem 18.5()
Problem: A 1.5 10-6 kg beetle has been knocked upside down. It arches his back, using its muscles to push against the floor and raise its center of mass so it can jump up in the air and flip over. Its center of mass is 0.25mm off the ground. The lowest it can get its center of mass is 0.15mm off the ground, and the highest it can get its center of mass before it can no longer push is 0.35mm. It can push against the floor with a force three times its own weight. (a)If you were to identify just the beetle as your system, what is the net external force on the beetle as it is pushing? Draw a free body diagram for the bug. (b)Assuming the beetle was at rest when it started pushing on the floor, what is its kinetic energy when its back leaves the floor? 7
(c)How much work did the floor do on the bug? Carefully draw a work-energy diagram showing how energy was transferred during the push for the defined system. (d)How high does its center of mass travel from the ground at the top of the jump? (You may want to change systems here.) Solution to Part (a) The bug's weight is the force that gravity exerts on it, or Fg = mg = 1.47 10-5 N. The bug can push at 3 times this force, or 4.41 10-5 N (so the ground pushes back on it with this force, Ff loor ). The earth is outside the system, so it exerts a force. The external force on the bug is the vector sum of all the forces acting on the bug. Since the forces are in opposite directions, the magnitude of the net external force on the bug is: Fext = (3 - 1)mg = 2.94 10-5 N
Ffloor
bug
Fg
Grading Key: Part (a) 5 Points 1 point(s) : Fg shown 1 point(s) : Ff loor > Fg 1 point(s) : Ff loor = 3Fg 1 point(s) : Ff loor opposite direction from Fg 1 point(s) : Net force consistent with diagram Solution to Part (b) The net force is exerted over a distance xcm = (3.5 10-4 m - 1.5 10-4 m) = 2 10-4 m, so the change in kinetic energy of the beetle while its center of mass moves through this distance is Kcm = Fext xcm = (2.94 10-5 N)(2 10-4 m) = 5.88 10-9 J Grading Key: Part (b) 4 Points 1 point(s) : Correct relationship 1 point(s) : correct xcm 1 point(s) : Fext from part (a) 1 point(s) : Correct answer with units. Can get full points if they do it correctly the hard way. Solution to Part (c) 8
0
U
K Esource
Work
Grading Key: Part (c) 8 Points 2 point(s) : Floor does no work on the bug 2 point(s) : Esource = K + W (work can be zero) 2 point(s) : No potential energy 1 point(s) : Extra credit: if Work negative
1 1 point(s) : Extra credit: if Work = 2 K in size (only if work is negative).
Solution to Part (d) When all of this kinetic energy has been converted to potential, it will stop moving, so put the earth back in the system and do conservation of energy: Kcm = Ucm = mgh and so h= (5.88 10-9 J) -4 m m = 4 10 (1.5 10-6 kg)(9.8 s2 )
This is the distance that its center of mass moves after its back leaves the ground (when its center of mass is 0.35mm high) so the total distance from the ground to the bug's center of mass is 4 10-4 m + 3.5 10-4 m = 7.5 10-4 m Grading Key: Part (d) 5 Points 2 point(s) : correct relationship 1 point(s) : initial K from (b) 1 point(s) : Remembers to add xcm initial 1 point(s) : uses xcm initial for height the bug leaves the floor. Total Points for Problem: 22 Points
9