D-sol - Chapter D - Problems Blinn College - Physics 2425 -...

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Chapter D - Problems Blinn College - Physics 2425 - Terry Honan Problem D.1 The same force acts at different times on two masses. m 1 is given an acceleration of 4 m ë s 2 and m 2 is given an acceleration of 1 m ë s 2 . (a) What is the ratio of the masses, m 2 ê m 1 ? (b) Suppose the two masses are combined and the same force is acted on the combination. What is its acceleration? Solution to D.1 (a) F = m 1 a 1 = m 2 a 2 , a 1 = 4 m s 2 and a 2 = 1 m s 2 m 2 m 1 = a 1 a 2 = 4 1 = 4 (a) F = m 1 a 1 = H m 1 + m 2 L a 3 , a 1 = 4 m s 2 and a 2 = 1 m s 2 m 1 a 1 = H m 1 + m 2 L a 3 and m 2 = 4 m 1 ï m 1 4 = H m 1 + 4 m 1 L a 3 ï a 3 = 4 5 m s 2 Problem D.2 A 2 ton truck provides an acceleration of 3 ft ë s 2 to a 4 ton trailer. If the truck exerts the same force on the road while pulling a 16 ton trailer, what acceleration results? Assume there are no resistive forces. Solution to D.2 Like pounds, tons are units of weight. lb-mass is often used to refer to the mass equivalent of 1 lb, so we will use the term ton-mass to refer to its mass equivalent. F = m 1 a 1 = m 2 a 2 ï a 2 a 1 = m 1 m 2 ï a 2 3 = 2 + 4 2 + 16 ï a 2 = 1 ft s 2 Note that the ratio of two masses in ton-mass is the same as in kg and the ratio of two accelerations in ft ë s 2 is the same as in m ë s 2 . Problem D.3 A 5 g bullet is accelerated to 300 m ê s down a 72 cm gun barrel. Assuming a uniform acceleration, what is the net force on the bullet in the gun barrel? Solution to D.3
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Use one dimensional kinematics to find the acceleration. v 0 = 0, v = 300 m s and D x = 0.72 m v 2 - v 0 2 = 2 a D x ï a = 300 2 2 μ 0.72 = 62 500 N Newton's second law gives the net force. F net = m a = 0.005 μ 62 500 = 312.5 N Problem D.4 The displacement of 20 kg mass as a function of time in SI units is given by: r H t L = Y 2 t 3 - 9 t - 15, - 6 t 2 + 5 t + 18 ] . What is the net force acting on the mass as a function of time? What is the net force at t = 2 s? Solution to D.4 The acceleration is the second derivative of the position vector. Newton's second law then gives the net force. v H t L = t r = t Y 2 t 3 - 9 t - 15, - 6 t 2 + 5 t + 18 ] = Y 6 t 2 - 9, - 12 t + 5 ] a H t L = t v = t Y 6 t 2 - 9, - 12 t + 5 ] = X 12 t , - 12 \ At t = 2 s we get a H 2 L = X 24, - 12 \ m s 2 ï F net = m a = 20 a = X 480, - 240 \ N. Problem D.5
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D-sol - Chapter D - Problems Blinn College - Physics 2425 -...

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