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Problem 26 Easy Difficulty

Review Conceptual Example 7 in preparation for this problem. In tests on earth a lunar surface exploration vehicle (mass $=5.90 \times 10^{3} \mathrm{kg}$ ) achieves a forward acceleration of 0.220 $\mathrm{m} / \mathrm{s}^{2} .$ To achieve this same acceleration on the moon, the vehicle's engines must produce a drive force of $1.43 \times 10^{3} \mathrm{N}$ . What is the magnitude of the frictional force that acts on the vehicle on the moon?

Answer

130 $\mathrm{N}$

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Video Transcript

To solve this question, you have to apply Newton's second law on the situation on the room. In the situation, there are two forces acting on the vehicle, the driving force on the frictional force. So Newton's second law tells us that the net force acting on the vehicle is ICO streets mass. Multiply it by its acceleration. No, the net force is composed by these two forces, the driving force and the frictional force. Let us set our reference point as pointing to the right. So everything that is pointing to the right will be positive and everything that is pointing to the left will be negative. As a consequence, then the driving force is positive and the frictional force is negative and we have mass times acceleration on the right side of the equation. Now we can plug in the values that were given by the problem. So we have 1.43 time stand to the third as a driving force. Miners there no frictional forces being equals to the mass off the eagle, 5.9 times 10 to the 30 kilograms times declaration off 0.22 meters per second squared. Then we can solve this equation for the driving force. To get that the driving forces are equal to 1.43 times 10 to the third miners, 5.9 thanks tend to the third time, 0.22 and these gives us a driving force of approximately 130 new times.