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A $1.14 \times 10^{4}$ -kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is 1.60 $\mathrm{m} / \mathrm{s}^{2}$ . At an altitude of 165 $\mathrm{m}$ the craft's downward velocity is 18.0 $\mathrm{m} / \mathrm{s}$ . To slow down the craft, a retrorocket is firing to provide an upward thrust. Assuming the descent is vertical, find the magnitude of the thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface.

29400 $\mathrm{N}$

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University of Michigan - Ann Arbor

Numerade Educator

Hope College

University of Winnipeg

to solve this question. We begin by discovering worries the neccessary acceleration to reduce the velocity of the craft to zero when it just this surface off the moon. For that we can you start a challenge equation. We tells us the following Velocity squared is given by the initial velocity squared, plus two times the acceleration times displacement. Let me choose my reference frame as being this vertical axis, pointing upwards. Then we apply Torricelli's equation as follows. The velocity squared should be close to zero at the end off the landing. The initial velocity waas 18. Then we have true times the acceleration which we don't know times the displacement. The displacement is miners 165 meters because displacement is happening on the negative direction on my reference frame. So times miners 165. Then we have to solve this equation for the acceleration and it goes as follows True times A times 165 Is it close to 18 squared, then eight is 18 squared, divided by true times 165. These results in an acceleration off approximately 0.982 meters per second squared notice the plus sign off the acceleration. It means that the acceleration points to this direction and then it's correct because the velocities downwards. So in order to reduce the velocity, the acceleration most point upwards. Now we have to complete one. Is the trust neccessary to produce such acceleration? For that, we use Newton's second law, which tells us that the net force is equal to the mass times acceleration off the craft. The Net force is composed by true forces, the trust force pointing upwards minus the weight force pointing downwards, and these easy coz the mass off the craft times its acceleration. Then the trust force is given by the weight force plus M times say the weight force is given by the mass off the craft times the surface gravity. Be careful because now we are at the moon, so we should use the surface gravity off the moon using the values that were given by the problem gives us a trust force that is given by 1.14 times 10 to the four times 1.6 plus 1.14 times stand to the fore Times 0.982 and these results in a trust force off approximately 29,000 and 400 mutants is the answer for this question.

Brazilian Center for Research in Physics