University Physics Volume 1
18th Edition
ISBN: 9781938168277
Author: William Moebs, Samuel J. Ling, Jeff Sanny
Publisher: OpenStax - Rice University
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Chapter 8, Problem 8.6CYU
Check Your Understanding Fend the forces on the particle in Example 8.6 when kinetic energy is 1.0 J at
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Chapter 8 Solutions
University Physics Volume 1
Ch. 8 - Check Your understanding In Example 8.1 what are...Ch. 8 - Check Your Understanding What are the values of...Ch. 8 - Check Your Understanding When the length of the...Ch. 8 - Check Your Understanding Suppose the mass in...Ch. 8 - Check Your Understanding A two-dimensional,...Ch. 8 - Check Your Understanding Fend the forces on the...Ch. 8 - Check Your Understanding How high above the bottom...Ch. 8 - Check Your Understanding You probably recall that,...Ch. 8 - Check Your Understanding What potential energy...Ch. 8 - Check Your Understanding Repeat Example 8.10 when...
Ch. 8 - Check Your Understanding Find x(t) for the...Ch. 8 - The kinetic energy of a system must always be...Ch. 8 - The force exerted by a diving board is...Ch. 8 - Describe the gravitational potential energy...Ch. 8 - A couple of soccer balls of equal mass are kiched...Ch. 8 - What is the dominant factor that affects the speed...Ch. 8 - Two people observe a leaf falling from a tree. One...Ch. 8 - What is the physical meaning of a non-conservative...Ch. 8 - A bottle rocket is shot straight up in the air...Ch. 8 - An external force acts on a particle during a trip...Ch. 8 - When a body slides down an inclined plane, does...Ch. 8 - Consider the following scenario. A car for which...Ch. 8 - A dropped ball bounces to one-half its original...Ch. 8 - “ E=K+Uconstant is a special case of the work...Ch. 8 - In a common physics demonstration, a bowling ball...Ch. 8 - A child jumps tip and down on a bed, reaching a...Ch. 8 - Can a non-conservative force increase the...Ch. 8 - Neglecting air resistance, how much would I have...Ch. 8 - A box is dropped onto a spring at its equilibrium...Ch. 8 - Using values from Table 8.1, how many DNA...Ch. 8 - If the energy in fusion bombs were used to supply...Ch. 8 - A camera weighing 10 N falls from a small drone...Ch. 8 - Someone drops a 50 — g pebble off of a docked...Ch. 8 - A cat’s crinkle ball toy of mass 15 g is thrown...Ch. 8 - A force F(x)=(3.0/x)N acts on a particle as it...Ch. 8 - A force F(x)=(5.0x2+7.0x)N acts on a particle as...Ch. 8 - Find the force corresponding to the potential...Ch. 8 - The potential energy function for either one of...Ch. 8 - A particle of mass 2.0 kg moves under the...Ch. 8 - A particle of mass 2.0 kg moves under the...Ch. 8 - A crate on rollers is being pushed without...Ch. 8 - A boy throws a ball of mass 0.25 kg straight...Ch. 8 - A mouse of mass 200 g falls 100 m down a vertical...Ch. 8 - Using energy considerations and assuming...Ch. 8 - A 1.0-kg ball at the end of a 2.0-m string swings...Ch. 8 - Ignoring details associated with friction, extra...Ch. 8 - Tarzan grabs a vine hanging vertically from a tall...Ch. 8 - Assume that the force of a bow on an arrow behaves...Ch. 8 - A 100 — kg man is skiing across level ground at a...Ch. 8 - A sled of mass 70 kg starts from rest and slides...Ch. 8 - A girl on a skateboard (total mass of 40 kg) is...Ch. 8 - A baseball of mass 0.25 kg is hit at home plate...Ch. 8 - A small block of mass in slides without friction...Ch. 8 - The massless spring of a spring gun has a force...Ch. 8 - A small ball is tied to a string and set rotating...Ch. 8 - A mysterious constant force of 10 N acts...Ch. 8 - A single force F(x)=4.0x (in newtons) acts on a...Ch. 8 - A particle of mass 4.0 kg is constrained to move...Ch. 8 - The force on a particle of mass 2.0 kg varies with...Ch. 8 - A 4.0-kg particle moving along the x -axis is...Ch. 8 - A particle of mass 0.50 kg moves along the x -axis...Ch. 8 - (a) Sketch a graph of the potential energy...Ch. 8 - In the cartoon movie Pocahontas...Ch. 8 - In the reality television show “Amazing Race”...Ch. 8 - In the Back to the Future movies...Ch. 8 - In the Hunger Games movie...Ch. 8 - In a “Top Fail” video...Ch. 8 - In a Coyote/Road Runner cartoon clip...Ch. 8 - In an iconic movie scene, Forrest Gump...Ch. 8 - In the movie Monty Python and the Holy Grail...Ch. 8 - A 60.0-kg skier with an initial speed of 12.0 m/s...Ch. 8 - (a) How high a hill can a car coast up (engines...Ch. 8 - A 5.00105kg subway train is brought to a stop from...Ch. 8 - A pogo stick has a spring with a spring constant...Ch. 8 - A block of mass 500 g is attached to a spring of...Ch. 8 - A block of mass 200 g is attached at the end of a...Ch. 8 - A T-shirt cannon launches a shirt at 5.00 m/s from...Ch. 8 - A child (32 kg) jumps up and down on a trampoline....Ch. 8 - Shown below is a box of mass m1 that sits on a...Ch. 8 - A massless spring with force constant k=200N/m...Ch. 8 - A particle of mass 2.0 kg moves under the...Ch. 8 - Block 2 shown below slides along a frictionless...Ch. 8 - A body of mass m and negligible size starts from...Ch. 8 - A mysterious force acts on all particles along a...Ch. 8 - An object of mass 10 kg is released at point A,...Ch. 8 - Shown below is a small ball of mass m attached to...Ch. 8 - A block leaves a frictionless inclined surface...Ch. 8 - A block of mass m, after sliding down a...Ch. 8 - A block of mass 300 g is attached to a spring of...Ch. 8 - Consider a block of mass 0.200 kg attached to a...Ch. 8 - A skier starts from rest and slides downhill. What...Ch. 8 - Repeat the preceding problem, but this time,...Ch. 8 - Two bodies are interacting by a conservative force...Ch. 8 - In an amusement park, a car rolls in a track as...Ch. 8 - A 200-g steel ball is tied to a 2.00m “massless”...Ch. 8 - A 300 g hockey puck is shot across an ice-covered...Ch. 8 - A projectile of mass 2 kg is fired with a speed of...Ch. 8 - An artillery shell is fired at a target 200 m...Ch. 8 - How much energy is lost to a dissipative drag...Ch. 8 - A box slides on a frictionless surface with a...
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- A particle moves along a curved path y(x)=(10m){1+cos[0.1m1]x} , from x=0 to x=10m , subject to a tangential force of variable magnitude F(x)=(10N)sin[(0.1m-1)x] . How much work does the force do? (Hint: Consult a table of integrals or use a numerical integration program.)arrow_forwardA particle moves in the xy plane (Fig. P9.30) from the origin to a point having coordinates x = 7.00 m and y = 4.00 m under the influence of a force given by F=3y2+x. a. What is the work done on the particle by the force F if it moves along path 1 (shown in red)? b. What is the work done on the particle by the force F if it moves along path 2 (shown in blue)? c. What is the work done on the particle by the force F if it moves along path 3 (shown in green)? d. Is the force F conservative or nonconservative? Explain. FIGURE P9.30 In each case, the work is found using the integral of Fdr along the path (Equation 9.21). W=rtrfFdr=rtrf(Fxdx+Fydy+Fzdz) (a) The work done along path 1, we first need to integrate along dr=dxi from (0,0) to (7,0) and then along dr=dyj from (7,0) to (7,4): W1=x=0;y=0x=7;y=0(3y2i+xj)(dxi)+x=7;y=0x=7;y=4(3y2i+xj)(dyj) Performing the dot products, we get W1=x=0;y=0x=7;y=03y2dx+x=7;y=0x=7;y=4xdy Along the first part of this path, y = 0 therefore the first integral equals zero. For the second integral, x is constant and can be pulled out of the integral, and we can evaluate dy. W1=0+x=7;y=0x=7;y=4xdy=xy|x=7;y=0x=7;y=4=28J (b) The work done along path 2 is along dr=dyj from (0,0) to (0,4) and then along dr=dxi from (0,4) to (7,4): W2=x=0;y=0x=0;y=4(3y2i+xj)(dyj)+x=0;y=4x=7;y=4(3y2i+xj)(dyi) Performing the dot product, we get: W2=x=0;y=0x=0;y=4xdy+x=0;y=4x=7;y=43y2dx Along the first part of this path, x = 0. Therefore, the first integral equals zero. For the second integral, y is constant and can be pulled out of the integral, and we can evaluate dx. W2=0+3y2x|x=0;y=4x=7;y=4=336J (c) To find the work along the third path, we first write the expression for the work integral. W=rtrfFdr=rtrf(Fxdx+Fydy+Fzdz)W=rtrf(3y2dx+xdy)(1) At first glance, this appears quite simple, but we cant integrate xdy=xy like we might have above because the value of x changes as we vary y (i.e., x is a function of y.) [In parts (a) and (b), on a straight horizontal or vertical line, only x or y changes]. One approach is to parameterize both x and y as a function of another variable, say t, and write each integral in terms of only x or y. Constraining dr to be along the desired line, we can relate dx and dy: tan=dydxdy=tandxanddx=dytan(2) Now, use equation (2) in (1) to express each integral in terms of only one variable. W=x=0;y=0x=7;y=43y2dx+x=0;y=0x=7;y=4xdyW=y=0y=43y2dytan+x=0x=7xtandx We can determine the tangent of the angle, which is constant (the angle is the angle of the line with respect to the horizontal). tan=4.007.00=0.570 Insert the value of the tangent and solve the integrals. W=30.570y33|y=0y=4+0.570x22|x=0x=7W=112+14=126J (d) Since the work done is not path-independent, this is non-conservative force. Figure P9.30ANSarrow_forwardCheck Your Understanding There is a second solution to the system of equations solved in this example (because the energy equation is quadratic): v1.f=-2.5m/s , v2.f=0 . This solution is unacceptable on physical grounds; what’s with it?arrow_forward
- A small block of mass m = 200 g is released from rest at point along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P8.43). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point relative to point . (b) the kinetic energy of the block at point . (c) its speed at point B, and (d) its kinetic energy and the potential energy when the block is at point . Figure P8.43 Problems 43 and 44.arrow_forwardTwo bodies are interacting by a conservative force Show that the mechanical energy of an isolated system consisting of two bodies interacting with a conservative force is conserved. (Hint: Start by using Newton’s third law and the definition of work to find the work done on each body by the conservative force.)arrow_forwardA nonconstant force is exerted on a particle as it moves in the positive direction along the x axis. Figure P9.26 shows a graph of this force Fx versus the particles position x. Find the work done by this force on the particle as the particle moves as follows. a. From xi = 0 to xf = 10.0 m b. From xi = 10.0 to xf = 20.0 m c. From xi = 0 to xf = 20.0 m FIGURE P9.26 Problems 26 and 27.arrow_forward
- A force F = (6i 2j) N acts on a panicle that under-goes a displacement r = (3i + j) m. Find (a) the work done by the force on the particle and (b) the angle between F and r.arrow_forwardConsider a particle on which a force acts that depends on the position of the particle. This force is given by . Find the work done by this force when the particle moves from the origin to a point 5 meters to the right on the x-axis.arrow_forwardA person in good physical condition can put out 100 W of useful power for several hours at a stretch, perhaps by pedaling a mechanism that drives an electric generator. Neglecting any problems of generator efficiency and practical considerations such as resting time: (a) How many people ‘would it take to nm a 4.O0-kW electric clothes dryer? (b) How many people would it take to replace a large electric power plant that generates 800 MW?arrow_forward
- A mysterious force acts on all particles along a particular line and always points towards a particular point P on the line. The magnitude of the force on a particle increases as the cube of the distance from that point; that is Fr3 , if the distance from P to the position of the particle is r. Let b be the proportionality constant, and write the magnitude of the force as F=br3. Find the potential energy of a particle subjected to this force when the particle is at a distance D from P, assuming the potential energy to be zero when the particle is at P.arrow_forwardIf the dot product of two vectors vanishes, what can you say about their directions?arrow_forwardA particle is subject to a force Fx that varies with position as shown in Figure P7.9. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d) What is the total work done by the force over the distance x = 0 to x = 15.0 m?arrow_forward
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