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SECTION 7.5 • Kinetic Energy and the Work—Kinetic Energy Theorem 193 7.5 Kinetic Energy and the Work–Kinetic Energy Theorem We have investigated work and identified it as a mechanism for transferring energy Consider a system consisting of a single object. Figure 7.13 shows a block of mass m moving through a displacement directed to the right under the action of a net force F, also directed to the right. We know from Newton’s second law that the block moves with an acceleration a. If the block moves through a displacement "r # "xˆi # (x f & x i )ˆ i , the work done by the net force F is (7.13) Using Newton’s second law, we can substitute for the magnitude of the net force + F # ma, and then perform the following chain-rule manipulations on the integrand: # W # $ x f x i
# Fdx # # v f ΣF m v i ∆ x Figure 7.13 An object undergoing a displacement "r # "x ˆi and a change in velocity under the action of a constant net force F. # F s mg d (c) (b) (a) Figure 7.12 (Example 7.6) Determining the force constant k of a spring. The elongation d is caused by the attached object, which has a weight mg. Because the spring force balances the gravitational force, it follows that k # mg/d. Example 7.6 Measuring k for a Spring A common technique used to measure the force constant of (A) If a spring is stretched 2.0 cm by a suspended object hav- ing a mass of 0.55 kg, what is the force constant of the spring? Solution Because the object (the system) is at rest, the up- g. In this case, we apply Hooke’s law to give 'F s ' # kd # mg, or (B) How much work is done by the spring as it stretches through this distance? 2.7 * 10 2 N/m k # mg d # (0.55 kg)(9.80 m/s 2 ) 2.0 * 10 & 2 m # Solution Using Equation 7.11, What If? Suppose this measurement is made on an eleva- tor with an upward vertical acceleration a. Will the unaware ex- Answer The force F s in Figure 7.12 must be larger than m g to produce an upward acceleration of the object. Because F s must increase in magnitude, and 'F s ' # kd, the spring must extend farther. The experimenter sees a larger extension for a # 0. Newton’s second law applied to the hanging object gives where k is the actual spring constant. Now, the experimenter 'F s ' # k-d # mg where k- is the spring constant as measured by the If the acceleration of the elevator is upward so that a y is posi- tive, this result shows that the measured spring constant will k- # mg d # mg % m(g ) a y ) k & # g g ) a y k d # m(g ) a y ) k kd & mg # ma y # F y # ' F s ' & mg # ma y & 5.4 * 10 & 2
J # W s
# 0 & 1 2 kd 2 # & 1 2 (2.7 * 10 2 N/m)(2.0 * 10 & 2 m) 2 |