3 Solution Set 3 on chapters 6&7

3 Solution Set 3 on chapters 6&7 - 1 Chapter 6:...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 6: Work and Energy . Selected Conceptual Questions, pp.187. Problems pp. 188: # 11, 26, 40, 44, 45, 49, 50, 57, 69, 86 Chapter 7: Impulse and Momentum . Selected Conceptual Questions, pp.216. Problems pp. 217: # 11, 18, 22, 23, 34, 42, 43, 45, 52 Chapter 6: Work and Energy 11. REASONING The work done by a constant force of magnitude F that makes an angle θ with a displacement of magnitude s is given by ( 29 cos W F s θ = (Equation 6.1). The magnitudes and directions of the pulling forces exerted by the husband and wife differ, but the displacement of the wagon is the same in both cases. Therefore, we can express the work W H done by the husband’s pulling force as ( 29 H H H cos W F s = . Similarly, the work done by the wife can be written as ( 29 W W W cos W F s = . We know the directional angles of both forces, the magnitude F H of the husband’s pulling force, and that both forces do the same amount of work ( W W = W H ). SOLUTION Setting the two expressions for work equal to one another, and solving for the magnitude F W of the wife’s pulling force, we find ( 29 W W cos F s ( 29 W H H cos W F s = 1 442 4 43 ( 29 ( 29 H H H W W 67 N cos58 cos or 45 N cos cos38 W F F = = = o o 1 442 4 43 26. REASONING a. The magnitude s of the displacement that occurs while the snowmobile coasts to a halt is the distance that we seek. The work W done by the net external force F acting on the snowmobile during the coasting phase is given by W = ( F cos θ ) s (Equation 6.1), and we can use this expression to obtain s . To do so, we will need to evaluate W . This we will do with the aid of the work-energy theorem as given by 2 2 1 1 2 2 f 0 W mv mv = - (Equation 6.3), since we know the mass m and the final and initial velocities v f and v 0 , respectively. The net external force F during the coasting phase is just 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
the horizontal force of kinetic friction (magnitude = f k ), since the drive force is shut off and the vertical forces balance (there is no vertical acceleration). This frictional force has a magnitude of 205 N. This follows because, while the snowmobile is moving at a constant 5.50 m/s, it is not accelerating, and the drive force must be balancing the frictional force, which exists both before and after the drive force is shut off. Finally, to use W = ( F cos θ ) s (Equation 6.1) to determine s , we will need a value for the angle θ between the frictional force and the displacement. The frictional force points opposite to the displacement, so θ = 180°. b. By using only the work-energy theorem it is not possible to determine the time t during which the snowmobile coasts to a halt. To determine t it is necessary to use the equations of kinematics. We can use these equations, if we assume that the acceleration during the coasting phase is constant. The acceleration is determined by the force of friction (assumed constant) and the mass of the snowmobile, according to Newton’s second law. For example, we can use ( 29 1 2 0 f
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/01/2011 for the course PHY 135 taught by Professor Wagihghobriel during the Spring '11 term at University of Toronto- Toronto.

Page1 / 15

3 Solution Set 3 on chapters 6&7 - 1 Chapter 6:...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online