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# hw6 - al-ameri(aoa434 Hwk06 Stokes(19102 This print-out...

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al-ameri (aoa434) – Hwk06 – Stokes – (19102) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is ONLINE HOMEWORK No 6. It is due before 7.00 am on Wednesday 3 March, Abu Dhabi Time (9.00 pm on Tuesday in Texas). 001 10.0 points The initial kinetic energy imparted to a 0 . 51 kg bullet is 1859 J. The acceleration of gravity is 9 . 81 m / s 2 . Neglecting air resistance, find the range of this projectile when it is fired at an angle such that the range equals the maximum height attained. Correct answer: 0 . 348882 km. Explanation: Let : m = 0 . 51 kg , K = 1859 J , and g = 9 . 81 m / s 2 . Assume that the firing height is negligible and that the bullet lands at the same elevation from which it was fired. The range R is given by R = v 2 0 g sin 2 θ . From kinematics, x = ( v 0 cos θ ) t and y = ( v 0 sin θ ) t 1 2 g t 2 . R = h max = ( v 0 sin θ ) 2 2 g , so v 2 0 g sin 2 θ = ( v 0 sin θ ) 2 2 g v 2 0 g (2 sin θ cos θ ) = ( v 0 sin θ ) 2 2 g tan θ = 4 θ = 76 . The kinetic energy is K = 1 2 m v 2 0 v 2 0 = 2 K m , so R = v 2 0 g sin 2 θ = 2 K m g sin 2 θ = 2 (1859 J) (0 . 51 kg) (9 . 81 m / s 2 ) sin 152 · km 1000 m = 0 . 348882 km . 002 (part 1 of 3) 10.0 points A 18 . 7 kg block is dragged over a rough, hor- izontal surface by a constant force of 85 . 8 N acting at an angle of 28 above the horizon- tal. The block is displaced 50 . 9 m, and the coefficient of kinetic friction is 0 . 19. The acceleration of gravity is 9 . 8 m / s 2 . 18 . 7 kg μ = 0 . 19 85 . 8 N 28 Find the work done by the 85 . 8 N force. Correct answer: 3856 . 03 J. Explanation: Let : m = 18 . 7 kg , F = 85 . 8 N , s x = 50 . 9 m , μ = 0 . 19 , and g = 9 . 8 m / s 2 . Consider the force diagram

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al-ameri (aoa434) – Hwk06 – Stokes – (19102) 2 F θ m g n f k Work is W = vector F · vectors , where vectors is the distance traveled. In this problem vectors = 5ˆ ı is only in the x direction. W F = F x s x = F s x cos θ = (85 . 8 N) (50 . 9 m) cos 28 = 3856 . 03 J . 003 (part 2 of 3) 10.0 points Find the work done by the force of friction. Correct answer: 1382 . 75 J. Explanation: To find the frictional force, F friction = μ N , we need to find N from vertical force balance. Note that N is in the same direction as the y component of F and opposite the force of gravity. Thus F sin θ + N = m g so that N = m g F sin θ . Thus the friction force is vector F friction = μ N ˆ ı = μ ( m g F sin θ ı . The work done by friction is then W μ = vector F friction · vectors = −| f μ | | s | = μ ( m g f μ sin θ ) s x = (0 . 19) [(18 . 7 kg) (9 . 8 m / s 2 ) (85 . 8 N) sin 28 ] (50 . 9 m) = 1382 . 75 J . 004 (part 3 of 3) 10.0 points If the block was originally at rest, determine its final speed.
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hw6 - al-ameri(aoa434 Hwk06 Stokes(19102 This print-out...

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