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# A small 8.00-kg rocket burns fuel that exerts a time- ISBN: 9780321675460 31

## Solution for problem 15E Chapter 4

University Physics | 13th Edition

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Problem 15E

A small 8.00-kg rocket burns fuel that exerts a time- varying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation F = A + Bt2. Measurements show that at t = 0, the force is 100.0 N, and at the end of the first 2.00 s, it is 150.0 N. (a) Find the constants A and B, including their SI units. (b) Find the ?net? force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after the fuel ignites. (c) Suppose that you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

Step-by-Step Solution:

Solution 15E Step 1 of 4: (a) Find the constants A and B, including their SI units. 2 F=A+Bt By substituting t= 0 and F=100 N in the given equation, 100 N= A+B(0) 2 A= 100 N By substituting A= 100 N, t= 2s and F= 150 N 2 150 N= 100N+B(2 s) 2 B= 12.5 N.s Using above results given equation becomes, F= 100+ 12.5t …………..1 Step 2 of 4: (b) Find the net force on this rocket and its acceleration (i) the instant after the fuel ignites Given data, Mass of rocket, m= 8 kg At the given instant refers to the time, t=0s so the force is F=100 N But for this force there is an resisting force due to the weight of the rocket, That is, F= m×g 2 F = 8 kg ×9.8m/s F= 78.4 N Therefore from newton's laws of motion, F net= m × a Force during motion - force against motion = m × a 100 N-78.4 N= 8 × a 21.6 N= 8 × a 2 a= 2.7m/s 2 Therefore, the acceleration of the rocket is 2.7m/s and net force is 21.6 N

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##### ISBN: 9780321675460

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