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In tests on earth a lunar surface exploration vehicle (mass = 5.86 × 103 kg) achieves a forward acceleration of 0.225 m/s2. To achieve this same acceleration on the moon, the vehicle's engines must produce a drive force of 1.42 × 103 N. What is the magnitude of the frictional force that acts on the vehicle on the moon?

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Answer:

[tex]f=101.5N[/tex]

Explanation:

The frictional force opposes the relative motion between the vehicle and the lunar surface. Thus, its sign is contrary to the drive force. According to Newton's second law:

[tex]\sum F=ma\\\sum F=F_d-f\\ma=F_d-f[/tex]

Rewriting for f:

[tex]f=F-ma\\f=1.42*10^3N-(5.86*10^3kg)(0.225\frac{m}{s^2})\\f=101.5N[/tex]

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