p0402

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Unformatted text preview: 194 CHAPTER 4. WORK AND ENERGY 4.2 Chapter 4, Problem 2 Problem: A block of mass m starts from rest at the top of an incline, slides a distance s and encounters a spring of constant k . The block compresses the spring by a displacement ∆s and then reverses direction. The coefficient of sliding friction on the incline is µ and the incline makes an angle θ with the horizontal. Determine the spring displacement, ∆s, as a function of k , s, θ, m, µ, and gravitational acceleration, g = |g|. Compute ∆s for s = 3.6 ft, µ = 0.25, k = 200 lb/ft, θ = 30o and mg = 10 lb. . . . .... . .... . .... . . . .. ............... . . ..................... . .. ....... .... ..... ............................ . ................................... . .... .... . .................................. . . .... . .. . . .................................. . ............................. .................................. . ... . . ................... . . .............................. . .............................. ................................. . ... .... . ........... . ..... . .. .... ................................... ........ .. .... .... . . . .... ... ...... ................................ ... . .............................. ........ ............. . . .. . ................... ........ ...... ... ......... ... ... ......................... ...... . . .......... ............ ... .. . ........................................... .... .. .... .. . .. . .. . ... .... ......... ..... .. .............................. . ............. ............. . ... .... ............. ... ...... . .. .... .... . ............................... .. ....... ........ ......... ... .... ... . .. ......... ................ .......... .................................... . .... .............................. .. ... .... ........................... .... ............................... ....... . .. ........... .... ................................ .............................. .... .. .... ... ..... ........ ............ .... . .... .......... .................. ........... . ............. .... ... .. . . .. ....... ...... ........... .... .............................. .... ............................ ..... ... ........ . .... . ... ..... .... ... ................................ ........ ..... ............ ..... .. .. .. .............................. ....... . ....... . .......... .................................. ..... . ...... . ................................ . ... . . .......................... ..... . .......... ........ .. . .... .... . .. ................................ ......... ....... .. ................. ..... . . . . ............ ............ .............. ....... ..... ... .............. . .... .. ................................. ..... . .. . ............... .... .... .... . .. ........... .. ................................ ... . ... . ................. ................. ...... .... . ................... ... .. .. . ... ... .. .... . ..... ..... .... .... ............................. ....... ...... .......... ...... ... ......... .... . ... ..... ..... . .. .... ................................. ....................... ....... ... ............. ... . .... ......... .... ...... ............. ...... ...................... .... ............ . . .. ............................. .. ... .... .. .. . .. .... ... . .... .... ........................................ .... . ... .. .... ... .. .. .. . . . .... . ...................................................... ..... . .... . ... . . ... ... ... ... .... .... ........... ................... . .. . . .............................. . . ....................... .. . . ................ . . .......................... ....... ...... ...... . . .. m g s k θ Solution: To solve, we use the Principle of Work and Energy for the box’s motion along the incline. Because it starts from rest and has zero velocity when the spring is compressed a distance ∆s, the initial and final values of the block’s kinetic energy, T1 and T2 , are both zero. The forces acting on the block when it’s on the incline are as shown below. Clearly, since the box is not accelerating in the direction normal to the chute, N = mg cos θ. Thus, the magnitude of the friction force is µmg cos θ, and it opposes the motion. The tangential component of the box’s weight accelerates it as it moves down the chute, and its magnitude is mg sin θ. Finally, the spring force at maximum compression is k ∆s. µN ........... . ... .... ........ .. . . .... .... ... ................... ................... .. .............. .................... .......................................... ..................... ......................................... ................ ................................... ... .. ..... .... ............... ...... .. ............... ................................. ........... ................................. .......... .... . ...................................... ............. .................. ................................ . .. ..... ...................... .... . ............. . .. ... . .......................... . ... .. .................... . ... ..... ... . .................. ... .......... ... .. ...... .. .................. .. .. . .......... .................... .. . .......... ....... .... .... .... . ............. .. .. . .. .. . .................... . . .......... . .................. . . ..................... .......... .. . .. .. . . ............. . ................ .... ........ . . . ................ . .. .............. ............. . ............... .. . . . ............ .............. .. . ............ . .............. . . ........ .. ..................... . ... . .. ............... ... . ..................... . ............ .. . .. . .................... .. ...... .. .. . .... . . .. . ............. .................................. .......... ....................... . k∆s N mg θ The work done by the three forces tangential to the surface is as it traverses the incline is s+∆s s+∆s U1−2 = 0 (mg sin θ − µmg cos θ)dx + s [−k (x − s)]dx 1 = mg (sin θ − µ cos θ) (s + ∆s) − k(∆s)2 2 The Principle of Work and Energy tells us that U1−2 = T2 − T1 Since T1 = T2 = 0, we have 1 mg (sin θ − µ cos θ) (s + ∆s) − k (∆s)2 = 0 2 Rearranging terms yields (∆s)2 + 2mg 2mg (µ cos θ − sin θ) ∆s + (µ cos θ − sin θ) s = 0 k k 4.2. CHAPTER 4, PROBLEM 2 Solving for ∆s, we have ∆s = mg (sin θ − µ cos θ) 1 ± k 1+ 2ks mg (sin θ − µ cos θ) 195 We are given s = 3.6 ft, µ = 0.25, k = 200 lb/ft, θ = 30o and mg = 10 lb. Hence, ∆s = 10 lb (sin 30o − 0.25 cos 30o ) 1 ± 200 lb/ft 1+ 2(200 lb/ft)(3.6 ft) (10 lb) (sin 30o − 0.25 cos 30o ) = (0.01415 ft)[1 ± 22.58] = −0.3054 ft and 0.3336 ft Clearly the negative root is not physically meaningful. Therefore, converting to inches, ∆s = 4.00 in ...
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