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# A spacecraft descends vertically near the surface of ## Problem 46P Chapter 4

University Physics | 13th Edition

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Problem 46P

A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0 kN from its engines slows it down at a rate of 1.20 m/s2, but it speeds up at a rate of 0.80 m/s2 with an upward thrust of 10.0 kN. (a) In each case, what is the direction of the acceleration of the spacecraft? (b) Draw a free-body diagram for the spacecraft. In each case, speeding up or slowing down, what is the direction of the net force on the spacecraft? (c) Apply Newton’s second law to each case, slowing down or speeding up, and use this to find the spacecraft’s weight near the surface of Planet X.

Step-by-Step Solution:

Solution 46P Step 1: Upward thrust F = 25 kN (Slows Down) 1 Upward thrust F = 10 kN (Speeds Up) 2 When slows down a = 1.2 m/s 2 1 When speeds up a = 0.8 m/s 2 2 Problem (a) To find the direction of the acceleration of the spacecraft Step 1: When the spacecraft slows down, The direction of the acceleration is upward because the spacecraft tries to land. Step 2: When the spacecraft speeds up, the direction of acceleration is downward. Problem (b) free -body diagram of net force Step 1: Slowing down, the net force is upward Program (c) To find the weight of the spacecraft Step 1: Mass of the body times the acceleration = net force When the spacecraft slows down ma =1F - W 1 ---- (1) When the spacecraft speeds up ma =2F - W 2 -----(2) Step 2: Dividing (1) by (2) a 1 F 1 W a = F W 2 2 Step 3: Simplifying the above equation, we get W = a1F 2a 2 1 a1 a 2 Step 4: Substituting values a 1nd a ar2 in opposite direction, a will h2ve negative sign 1.* 1* 1000(0.*) *5 1000 W = 1.2(0.8) 3 W = 1.6x10 N

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##### ISBN: 9780321675460

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