In tests on earth a lunar surface exploration vehicle (mass = 5.90 103 kg) achieves a forward acceleration of 0.215 m/s2. To achieve this same acceleration on the moon, the vehicle's engines must produce a drive force of 1.64 103 N. What is the magnitude of the frictional force that acts on the vehicle on the moon?
Fap - Ff =ma.
1640 - Ff = 5900*0.215 = 1269 N.
-Ff = 1269 - 1640 = -371
Ff = 371 N. = Force of friction.
Why did the lunar vehicle join a comedy club? Because it wanted to explore the moon's surface with a good laugh! But let me calculate the frictional force for you. We know that force is equal to mass multiplied by acceleration, so the force on the moon is 5.90 * 10^3 kg * 0.215 m/s^2 = 1.2685 * 10^3 N. Since the frictional force is equal to the force on the moon (since there's no need to overcome gravity), the magnitude of the frictional force is 1.2685 * 10^3 N.
To find the magnitude of the frictional force acting on the vehicle on the moon, we first need to understand the concept of friction and how it relates to the vehicle's acceleration.
Friction is a force that opposes the motion of objects in contact with each other. It depends on the nature of the surfaces in contact and the force pressing them together. In this case, the frictional force is responsible for decelerating the vehicle on the moon, opposing its forward motion.
According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration:
Net force = mass × acceleration
In this problem, the forward acceleration achieved on Earth is given as 0.215 m/s², and the mass of the vehicle is 5.90 × 10³ kg. Hence, the net force required to achieve this acceleration on Earth can be calculated as follows:
Net force on Earth = (5.90 × 10³ kg) × (0.215 m/s²)
Now, to achieve the same acceleration on the moon, the vehicle's engines must produce a drive force of 1.64 × 10³ N. We can set up an equation as follows:
Net force on Moon = Drive force - Frictional force
Since we are trying to find the magnitude of the frictional force, let's rearrange the equation:
Frictional force = Drive force - Net force on Moon
Frictional force = (1.64 × 10³ N) - Net force on Moon
Now, we know that the net force on Earth is equal to the net force on the moon. Therefore, the frictional force acting on the moon can be calculated as follows:
Frictional force on the moon = (1.64 × 10³ N) - Net force on Earth
Substituting the values, we get:
Frictional force on the moon = (1.64 × 10³ N) - [(5.90 × 10³ kg) × (0.215 m/s²)]
Simplifying this expression will give us the magnitude of the frictional force acting on the vehicle on the moon.
To find the magnitude of the frictional force acting on the vehicle on the moon, we can use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and its acceleration. In this case, the net force is the sum of the drive force and the frictional force. Therefore, the equation we can use is:
Net force = Drive force + Frictional force
Let's denote the frictional force as Ff.
On Earth:
Mass of the vehicle, m = 5.90 * 10^3 kg
Forward acceleration on Earth, a_e = 0.215 m/s^2
Using Newton's second law on Earth, we have:
Net force on Earth = m * a_e
On the Moon:
Drive force on the Moon, Fd_m = 1.64 * 10^3 N
Forward acceleration on the Moon, a_m = 0.215 m/s^2
Using Newton's second law on the Moon, we have:
Net force on the Moon = m * a_m
Since the magnitudes of the net forces on Earth and on the Moon are the same (as they both produce the same forward acceleration of 0.215 m/s^2), we can equate the two equations:
m * a_e = m * a_m + Ff
Simplifying the equation, we find:
Ff = m * (a_e - a_m)
Plugging in the given values:
Ff = 5.90 * 10^3 kg * (0.215 m/s^2 - 0.215 m/s^2)
Ff = 5.90 * 10^3 kg * (0)
Ff = 0 N
Therefore, the magnitude of the frictional force that acts on the vehicle on the moon is 0 N.