Weekly Test 6: Work-Energy, Power Begin Date: 3/18/2016 12:00:00 AM Due Date: 3/

Question

Weekly Test 6: Work-Energy, Power Begin Date: 3/18/2016 12:00:00 AM Due Date: 3/18/2016 11:59:00 PM End Date: 3/18/2016 11:59:00 PM (50%) Problem 2: As a 100-kg object falls toward the Earth in space, the gravitational force is directed toward the Earth and is modeled by the function Fr) k r2, where k is a constant equal to 4.0 x 1016 N.m? Otheexpertta.com G A 33% Part (a) Write an equation for the work done by gravity as a 100-kg object falls from a height RI above the earth to a height R2 Grade Summary Deductions 0% Potential 100 0 a b 7 8 9 HOME Submissions Attempts remaining: 3 4 5 6 (4% per attempt) 1 2 3 detailed view Hint Submit I give up! Hints 4% deduction per hint. Hints remaining: 3 Feedback 5% deduction per feedback. A 33% Part (b) Calculate the work done by gravity in megajoules (MJ) as the 100-kg object falls from a height R 42000 km to a height R2 8900 km. A 33% Part (c) Calculate the work done by gravity in megajoules (MJ) as the 100-kg object falls from a very large distance (infinity) to a height R 8900 km

Explanation / Answer

We know that,

Workdone = Force x distance

F(r) = -k/r2 ; k = 4 x 1016 N-m2

(a) Work done will be:

W = -k/R12 x R1 - ( - k/R22 x R2 ) = k/R2 - k/R1

W = k (1/R2 - 1/R1)

(b) R1 = 42000 km = 4.2 x 107 m ; R1 = 8900 km = 4.9 x 106 m

W = k (1/R2 - 1/R1) = 4 x 1016 ( 1/8.9 x 106 - 1/4.2 x 107 )

W = 4 x 1010 (1/8.9 - 1/42) = 3.54 x 109 J = 3540 M J

Hence, W = 3540 MJ

c) R1 = infinity ; R2 = 8900 km = 8.9 x 106 m

W = k (1/R2 - 1/R1) = 4 x 1016 ( 1/8.9 x 106 - 1/infintiy)

W = 4 x 1010 (1/8.9 - 0) = 4.5 x 109 J = 4500 M J

Hence, W = 4500 MJ

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