11. A block of mass mi 2.40 kg is connected to a second block of mass m2 = 1.80 kg, as shown in the sketch. When the blocks are released from rest, they move through a distance d = 0.500 m, at which point m₂ hits the floor. Given that the coefficient of kinetic friction between m₁ and the horizontal surface is μ = 0.450, find the speed of the blocks just before m₂ lands. A 1111 m, -0

11. A block of mass mi 2.40 kg is connected to a second block of mass m2 = 1.80 kg, as shown in the sketch. When the blocks are released from rest, they move through a distance d = 0.500 m, at which point m₂ hits the floor. Given that the coefficient of kinetic friction between m₁ and the horizontal surface is μ = 0.450, find the speed of the blocks just before m₂ lands. A 1111 m, -0

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter7: Conservation Of Energy

Section: Chapter Questions

Problem 79P: A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring...

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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.
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.
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
Jane, whose mass is 50.0 kg, needs to swing across a river (having width D) filled with person-eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force F, on a vine having length L and initially making an angle with the vertical (Fig. P7.81). Take D = 50.0 m, F = 110 N, L = 40.0 m, and = 50.0. (a) With what minimum speed must Jane begin her swing to just make it to the other side? (b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume Tarzan has a mass of 80.0 kg.
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A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (Fig. P7.79). The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point , the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The blocks speed at the bottom of the track is = 12.0 m/s, and the block experiences an average friction force of 7.00 N while sliding up the track. (a) What is x? (b) If the block were to reach the top of the track, what would be its speed at that point? (c) Does the block actually reach the top of the track, or does it fall off before reaching the top?
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A large cruise ship of mass 6.50 107 kg has a speed of 12.0 m/s at some instant. (a) What is the ships kinetic energy at this time? (b) How much work is required to stop it? (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of 2.50 km?
If the dot product of two vectors vanishes, what can you say about their directions?
Q12. The two tugboats each exert a constant force F = 20 N on the ship as shown in FigureQ12 These forces are always directed perpendicular to the ship's centerline. If the ship has amass m and a radius of gyration about its center of mass G of kG, determine the work done ofthe ship after it turns 80°. The ship is originally at rest. Where d = 6 m.
Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fills a pitcher full of Cola, places it on the counter top and gives the 2.6-kg pitcher a 8.8 N forward push over a distance of 48 cm to send it to a customer at the end of the counter. The coefficient of friction between the pitcher and the counter top is 0.28. a. Determine the work done by Pete on the pitcher during the 48 cm push.b. Determine the work done by friction upon the pitcher .c. Determine the total work done upon the pitcher .d. Determine the kinetic energy of the pitcher when Pete is done pushing it.e. Determine the speed of the pitcher when Pete is done pushing it.