Question

In tests on earth a lunar surface exploration vehicle (mass = 5.90 103 kg) achieves a forward acceleration of 0.230 m/s2. To achieve this same acceleration on the moon, the vehicle's engines must produce a drive force of 1.69 103 N. What is the magnitude of the frictional force that acts on the vehicle on the moon?

Answer 1

Frictional Force = 0.3337931 N

Step-by-step explanation:

Given

Mass of Vehicle = 5.90103kg

Acceleration on earth = 0.230m/s2

Drive force on Moon = 1.69103N

Remember that friction always opposes the direction of motion, safe to say that the frictional force will also act against the drive force. On the moon the drive force required to achieve an acceleration of 0.230m/s2 is 1.69103N but it is not the actual force being used for the acceleration, this total drive force also incorporates the force required to overcome friction.

The force required to achieve acceleration can be found using

`F=ma`

Where

F = Force

m = Mass

a = Acceleration

`F=(5.90103)(0.230)`

`F= 1.3572369`

This is the force required to produce acceleration

To find the frictional force we subtract this value from the total drive force

`F_(f)  = F_(d)  - F\\ F_(f)  = 1.69103  - 1.3572369\\ F_(f)  = 0.3337931 ‬`

Answer 2

Answer: 333N

Step-by-step explanation:

Using Newton’s second law,

1690 N - f = (5.9 x 10^3 ) x ( 0.23)

Therefore frictional force f =

f = 1690 - (5900 x 0.23)

f = 330N