A system in which only one particle can move has the potential energy shown in (Figure 1). Suppose U₁ = 76 J. Part A What is the y-component of the force on the particle at y = 0.5 m ? Express your answer with the appropriate units. Figure U (J) U₁ 0 y (m) 0 1 2 3 4 5 ☐ μÅ Fy = Value Units Submit Request Answer Part B ? 1 of 1 What is the y-component of the force on the particle at y = 4 m? Express your answer with the appropriate units. μÅ ? Fy = Value Units Submit Request Answer Provide Feedback

A system in which only one particle can move has the potential energy shown in (Figure 1). Suppose U₁ = 76 J. Part A What is the y-component of the force on the particle at y = 0.5 m ? Express your answer with the appropriate units. Figure U (J) U₁ 0 y (m) 0 1 2 3 4 5 ☐ μÅ Fy = Value Units Submit Request Answer Part B ? 1 of 1 What is the y-component of the force on the particle at y = 4 m? Express your answer with the appropriate units. μÅ ? Fy = Value Units Submit Request Answer Provide Feedback

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter8: Potential Energy And Conservation Of Energy

Section: Chapter Questions

Problem 25P: A force F(x)=(5.0x2+7.0x)N acts on a particle as it moves along the positive x-axis. (a)How much...

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A 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?
A horizontal force of 20 N is required to keep a 5.0 kg box traveling at a constant speed up a frictionless incline for a vertical height change of 3.0 m. (a) What Is the work done by gravity dining this change in height? (b) What Is the work done by the normal force? (c) What is the work done by the horizontal farce?
(a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1hp=746W) to reach a speed of 15.0 m/s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m-high hill in the process?
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.
A particle moves along the xaxis from x = 12.8 m to x = 23.7 m under the influence of a force F=375x3+3.75x where F is in newtons and x is in meters. Using numerical integration, determine the work done by this force on the particle during this displacement. Your result should he accurate to within 2%.
A block of mass 200 g is attached at the end of a massless spring of spring constant 50 N/m. The other end of the spring is attached to the ceiling and the mass is released at a height considered to be where the gravitational potential energy is zero. (a) What is the net potential energy of the block at the instant the block is at the lowest point? (b) What is the net potential energy of the block at the midpoint of its descent? (c) What is the speed of the block at the midpoint of its descent?
The force acting on a particle is Fx = (8x 16), where F is in newtons anti x is in meters. (a) Make a plot of this force versus x from x = 0 to x = 3.00 m. (b) From your graph, find the net work done by this force on the particle as it moves from x = 0 to x = 3.00 m.
A 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 P9.30ANS
Integrated Concepts A 75.0-kg cross-country skier is climbing a 3.0° slope at a constant speed of 2.00 m/s and encounters air resistance of 25.0 N. Find his power output for work done against the gravitational force and air resistance. (b) What average force does he exert backward on the snow to accomplish this? (c) If he continues to exert this force and to experience the same air resistance when he reaches a level area, how long will it take him to reach a velocity of 10.0 m/s?
If the net work done by external forces on a particle is zero, which of the following statements about the particle must be true? (a) Its velocity is zero. (b) Its velocity is decreased. (c) Its velocity is unchanged. (d) Its speed is unchanged. (e) More information is needed.
An object of mass m = 5.8 kg moves under the influence of one force. That force causes the object to move along a path given by x = 6.0 + 5.0t + 2.0t2, where x is in meters and t is in seconds. Calculate the work done by the force on the object from t = 2.0 s to t = 7.0 s.
A force F=(4xi+3yj), where F is in newtons and x and y are in meters, acts on an object as the object moves in the x direction from the origin to x = 5.00 m. Find the work W=Fdr done by the force on the object.