chap 4 - FORCE AND MOTION 4 EXERCISES Section 4.2 Newtons...

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4.1 FORCE AND MOTION EXERCISES Section 4.2 Newton’s First and Second Laws 13. INTERPRET We interpret this as a problem involving the application of Newton’s second law. The object under consideration is the train and the physical quantity of interest is the net force acting on it. DEVELOP The net force can be found by using Equation 4.3, r Fm a net = . EVALUATE Using Equation 4.3, the magnitude of the force acting on the train is found to be a net kg m s N == × = × (. ) / ) . 15 10 25 375 10 62 6 ASSESS The result is reasonable, since by definition, one newton is the force required to accelerate a 1-kg mass at the rate of 1 m/s 2 . 14. Ignoring the probable presence of other forces, we can apply Equation 4.3 to find (a) aF m ==× /( . ) / 12 10 5 N ) . / , 61 10 197 42 ×= kg m s and (b) a × = ) / ) . / . 146 10 821 56 2 Nk g c m s 15. INTERPRET We interpret this as a problem involving the application of Newton’s second law. The object under consideration is the airplane and the physical quantity of interest is the plane’s mass. DEVELOP We shall 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 plane’s mass can be found by using Equation 4.3, r r a net = . EVALUATE Using Equation 4.3, the mass of the plane is found to be m F a × net N ms kg 11 10 72 153 10 4 2 3 . ./ . ASSESS First, the units are consistent since 11 N kg m/s 2 =⋅ . The result is reasonable, since by definition, one newton is the force required to accelerate a 1-kg mass at the rate of 1 m/s 2 . 16. 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 av t av =− () / , 0 0 and the average net force on the passenger, while coming to rest, is a m v t av av kg m s −= 0 60 110 3 6 / ( )( / . )( / )/ ) . 014 131 sk N , 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.) 17. INTERPRET In this problem we want to find the relationship between the initial speed of the car and the force required to stop it. DEVELOP From Equation 4.3, we see that the net force on a car of given mass is proportional to the acceleration, Fa net ; . We can then relate the three quantities, displacement, velocity, and acceleration, by Equation 2.11, vv a x x 2 0 2 0 2 =+ . EVALUATE To stop a car in a given distance, , xx 0 the acceleration is a vv v 2 0 2 0 0 2 0 22 () () . Therefore, we see that Fv net ; 0 2 . Doubling v 0 quadruples the magnitude of F net .
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chap 4 - FORCE AND MOTION 4 EXERCISES Section 4.2 Newtons...

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