chap5 - CHAPTER 5 FORCE AND MOTION ActivPhysics can help...

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Unformatted text preview: CHAPTER 5 FORCE AND MOTION ActivPhysics can help with these problems: All activities in Section 2, Forces and Motion Section 5-4: Newton’s Second Law Problem 1. A subway train has a mass of 1 . 5 × 10 6 kg . What force is required to accelerate the train at 2 . 5 m/s 2 ? Solution F = ma = (1 . 5 × 10 6 kg)(2 . 5 m/s 2 ) = 3 . 75 MN. (This is the magnitude of the net force acting; see Table 1-1 for SI prefixes.) Problem 2. A railroad locomotive with a mass of 6 . 1 × 10 4 kg can exert a force of 1 . 2 × 10 5 N. At what rate can it accelerate (a) by itself and (b) when pulling a 1 . 4 × 10 6-kg train? Solution Ignoring the probable presence of other forces, we can apply Equation 5-3 to find (a) a = F/m = (1 . 2 × 10 5 N) / (6 . 1 × 10 4 kg) = 1 . 97 m/s 2 , and (b) a = (1 . 2 × 10 5 N) / (1 . 46 × 10 6 kg) = 8 . 21 cm/s 2 . Problem 3. A small plane starts down the runway with acceleration 7 . 2 m/s 2 . If the force provided by its engine is 1 . 1 × 10 4 N , what is the plane’s mass? Solution If we assume that the runway is horizontal (so that the vertical force of gravity and the normal force of the surface cancel) and neglect aerodynamic forces (which are small just after the plane begins to move) then the net force equals the engine’s thrust and is parallel to the acceleration. The horizontal component of Equation 5-3 gives the airplane’s mass, m = F/a = (1 . 1 × 10 4 N) / (7 . 2 m/s 2 ) = 1 . 53 × 10 3 kg. Problem 4. A car leaves the road traveling at 110 km/h and hits a tree, coming to a complete stop in 0.14 s. What average force does a seatbelt exert on a 60-kg passenger during this collision? Solution Assume that the seatbelt holds the passenger firmly to the seat, so that the passenger also stops in 0.14 s without incurring any secondary impact. Then the passenger’s average acceleration is a av = (0 − v ) /t , and the average net force on the passenger, while coming to rest, is F av = ma av = − mv /t = − (60 kg)(110 / 3 . 6)(m/s) / (0 . 14 s) = − 13 . 1 kN, or about 1.5 tons. (Here, we used the one-dimensional form of Newton’s second law. The minus sign indicates that the force is opposite to the direction of the initial velocity. It is reasonable to assume that this is a component of the force exerted by the seatbelt.) Problem 5. In an x-ray tube, electrons are accelerated to speeds on the order of 10 8 m/s, then slammed into a target where they come to a stop in about 10 − 18 s. Estimate the average stopping force on each electron. Solution The magnitude of the average force is ¯ F = m ¯ a = m | Δ v/ Δ t | = (9 . 11 × 10 − 31 kg)(10 8 m/s) / (10 − 18 s) ≃ 9 × 10 − 5 N. Compared to the TV tube in Example 5-2, the electron in an x-ray tube experiences a force billions of times greater. It is a result of the violence of this interaction that x rays, called bremsstrahlung, are emitted (see Problem 2-45)....
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This note was uploaded on 10/31/2010 for the course PHYSICS 69054 taught by Professor Huang during the Fall '09 term at Berkeley.

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chap5 - CHAPTER 5 FORCE AND MOTION ActivPhysics can help...

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