Explanation Given: The mass of the rocket sled is 4 Mg . Initially the rocket sled is at rest. The given F − t graph is shown in Figure (1). Draw the free body diagram of the rocket sled as shown in Figure (2). Write the principle of linear impulse and momentum. m v 1 + ∫ t 1 t 2 T d t = m v 2 (I) Here, m is the mass of the rocket sled, v is the velocity of the rocket sled, T is the thrust acting on the rocket sled, t is the time and the suffixes 1 , 2 indicates the initial and final stages. Write the formula for displacement of rocket sled. v = d s d t d s = v d t (II) Here, v is the velocity and d s d t is the rate of change of displacement with respect to time. Conclusion: Calculate the impulse on therocket sled due to the thrust ( T ) at the time interval of t = 0 to 5 s . Refer figure (1). The equation of the force acting on the rocket sled at the time interval of t = 0 to 2 5 s is T = 4 t 1 / 2 × 10 3 The impulse on the rocket sled due to the thrust ( T ) is ∫ 0 t T d t . Substitute 4 t 1 / 2 × 10 3 for T . ∫ 0 t 4 t 1 / 2 × 10 3 d t = 4 t 1 / 2 × 10 3 [ 2 t 3 / 2 3 ] 0 t = 4 × 10 3 × 2 t 3 / 2 3 − 0 = 4 × 10 3 × 2 t 3 / 2 3 Here, the total impulse acting on the rocket sled is ∑ ∫ t 1 t 2 T d t = ∫ 0 t 4 t 1 / 2 × 10 3 d t = 4 × 10 3 × 2 t 3 / 2 3 . Calculate the velocity of the rocket sled at the time interval of t = 0 to 2 5s . Substitute 4 Mg for m , 0 for v 1 and 4 × 10 3 × 2 t 3 / 2 3 for ∑ ∫ t 1 t 2 T d t in Equation (I). 4 Mg ( 0 ) + ( 4 × 10 3 × 2 t 3 / 2 3 ) = 4 Mg × 10 3 kg 1 Mg ( v 2 ) 4000 v 2 = 4000 2 3 t 3 / 2 ( v 2 ) 0 − 25 s = 2 3 t 3 / 2 When, t = 25 s . Substitute 25 s for t . ( v 2 ) 0 − 25 s = 2 3 ( 25 ) 3 / 2 = 83.3333 m/s Calculate the impulse on the rocket sled due to the thrust ( T ) at the time interval of t = 25 s to 35 s . Refer figure (1). The equation of the force acting on the rocket sled at the time interval of t = 25 s to 35 s is T − ( 20 × 10 3 ) t − 25 = 0 − ( 20 × 10 3 ) 35 − 25 T − ( 20 × 10 3 ) = − 2 × 10 3 ( t − 25 ) T = − 2000 t + 50000 + 20000 T = − 2000 t + 70000 The impulse on the rocket sled due to the thrust ( T ) is ∫ 25 t T d t . Substitute T = − 2000 t + 70000 for F . ∫ 25 t ( − 2000 t + 70000 ) d t = [ − 2000 t 2 2 + 70000 t ] 25 t = [ − 1000 t 2 + 70000 t ] − [ − 1000 ( 25 2 ) + 70000 ( 25 ) ] = − 1000 t 2 + 70000 t + 625000 − 1750000 = − 1000 t 2 + 70000 t − 1125000 Here, the total impulse acting on the rocket sled is ∑ ∫ t 1 t 2 F d t = ∫ 5 t ( 50 t + 350 ) d t = − 1000 t 2 + 70000 t − 1125000

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

Videos

Question

Chapter 15.2, Problem 22P

To determine

The speed and displacement of the rocket sled at t=35 s .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 15.2, Problem 21P

Chapter 15.2, Problem 23P

Students have asked these similar questions

Packages having a mass of 4-kg are delivered from a conveyor to a ramp with a velocity of v = 0.8 m/s. What is the velocity in which the packages will leave the ramp (at point B)? Consider e =30° andh = 12 m. The coefficient of the kinetic friction between the package and the ramp is = 0.35 Note. Please write your final answer in m/s in the provided box. Report your answer in 2 decimal places. For example: 2.34 m/s 0.8 m/s

The 333-Mg jet airliner has three engines, each of which produces a nearly constant thrust of 230 kN during the takeoff roll. Determine the lengths of runway required if the takeoff speed is 213 km/h. Compute s first for an uphill takeoff direction from A to B and second for a downhill takeoff from B to A on the slightly inclined runway. Neglect air and rolling resistance. Assume 0 = 0.7°. 0 Answers: Horizontal Uphill from A to B: Downhill from B to A: S= i S= i B m 3

The 5.27 kg collar B rests on the frictionless arm AA! The collar is held in place by the rope attached to drum D and rotates about O in a horizontal plane. The linear velocity of the collar B is increasing according to v = 0.2 t2 where v is in m/s and tis in seconds. Find the tension in the rope and the force of the bar on .the collar if 5 s,r= 0.558 m and 0 = 58° A A' D

Chapter 15 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

Ch. 15.2 - Determine the impulse of the force for t = 2 s.Ch. 15.2 - The 0.5kg ball strikes the rough ground and...Ch. 15.2 - Prob. 2FP Ch. 15.2 - Prob. 3FP Ch. 15.2 - The wheels of the 1.5-Mg car generate the traction...Ch. 15.2 - Prob. 5FP Ch. 15.2 - A man kicks the 150-g ball such that it leaves the...Ch. 15.2 - If the coefficient of kinetic friction between the...Ch. 15.2 - Prob. 3P Ch. 15.2 - Each of the cables can sustain a maximum tension...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The speed and displacement of the rocket sled at t = 35 s .

Solution Summary: The author explains how to calculate the velocity and displacement of the rocket sled at t=35.

Concept explainers

Videos

The speed and displacement of the rocket sled at t=35 s .

Want to see the full answer?

Chapter 15 Solutions