Is positive or negative work done by a constant force
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- A small particle of mass m is pulled to the top of a friction less half-cylinder (of radius R) by a light cord that passes over the top of the cylinder as illustrated in Figure P7.15. (a) Assuming the particle moves at a constant speed, show that F = mg cos . Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times. (b) By directly integrating W=Fdr, find the work done in moving the particle at constant speed from the bottom to the top of the hall-cylinder. Figure P7.15arrow_forwardA force F=(6j2j)N acts on a particle that undergoes 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_forwardA block of mass m = 0.250 kg is pressed against a spring resting on the bottom of a plane inclined an angle = 45.0 to the horizontal. The spring, which has a force constant of 955 N/m, is compressed a distance of 8.00 cm, and the block is released from rest. Consider the total energy of the springblockEarth system. a. What is the total distance the block moves from its initial position if the incline is frictionless? b. What is the total distance the block moves from its initial position if the coefficient of kinetic friction between the incline and the block is 0.330?arrow_forward
- A system consists of three particles, each of mass 5.00 g, located at the corners of an equilateral triangle with sides of 30.0 cm. (a) Calculate the gravitational potential energy of the system. (b) Assume the particles are released simultaneously. Describe the subsequent motion of each. Will any collisions take place? Explain.arrow_forwardA 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%.arrow_forwardA fellow student tells you that she has both zero kinetic energy and zero potential energy. Is this possible? Explain.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_forward(a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.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_forward
- A bungee cord is essentially a very long rubber band that can stretch up to four times its unstretched length. However, its spring constant vanes over its stretch [see Menz, P.G. “The Physics of Bungee Jumping.” The Physics Teacher (November 1993) 31: 483-487]. Take the length of the cord to be along the direction and define the stretch as the length of the cord minus its un-stretched length that is, (see below). Suppose a particular bungee cord has a spring constant, for of and for. (Recall that the of (Recall that the spring constant is the slope of the force versus its stretch (a) What is the tension in the cord when the stretch is 16.7 m (the maximum desired for a given jump)? (b) How much work must be done against the elastic force of the bungee cord to stretch It 16.7 m? Figure 7.16 (credit modification of work by Graeme Churchard)arrow_forward(a) 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. (b) What If? Find the work W=Fdr done by the force on the object if it moves from the origin to (5.00 m, 5.00 m) along a straightline path making an angle of 45.0 with the positive x axis. Is the work done by this force dependent on the path taken between the initial and final points?arrow_forwardAn 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.arrow_forward
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