in a tug of war a jeep of a mass 1400kg and a tractor of a mass of 2000kg pull on a horizontal rope in opposite directions. at one instant, the tractor pulls on the rope with a force of 1.50*10^4N while its wheels push horizontally against the groud with a force of 1.60*10^4N Calculate the instanous accelerations of the tractor and of the jeep and also calculate the horizontal push of the wheels of the jeep on the ground.

Net force=ma

1.60E4-1.50E4=(1400+2000)a
solve for a. Remember, the vehicles are attached, both masses are moved by the same net force.

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A girl pulls a sled along a level dirt road by means of a rope attached to the front of the sled. The mass of the sled is 40kg the coefficient of friction is. 600, and the angle between the rope and the road is 30 degrees. What pull must the girl exert to move the sled at constant velocity?

hello

Well, well, well! It seems we have quite a tug of war going on between a jeep and a tractor. Let's put on our thinking caps and calculate the instantaneously comical accelerations, shall we?

First things first, we need to calculate the net force acting on both the tractor and the jeep. The net force is equal to the force applied by the tractor minus the force acting against it due to the jeep.

For the tractor:
Net force = Force applied by the tractor - Force of the wheels on the ground

Net force = 1.50 * 10^4 N - 1.60 * 10^4 N
Net force = -0.10 * 10^4 N, oh dear, negative force!

Now, let's move on to the jeep. Since it's getting pulled in the opposite direction, the net force can be calculated as:

Net force = Force of the wheels on the ground - Force applied by the tractor

Net force = 1.60 * 10^4 N - 1.50 * 10^4 N
Net force = 0.10 * 10^4 N, a positive force with a little twist!

Now, let's calculate the instantaneously hilarious acceleration using Newton's second law:

Acceleration = Net force / Mass

For the tractor:
Acceleration of the tractor = (-0.10 * 10^4 N) / (2000 kg)
Acceleration of the tractor = -0.05 m/s², a snail's pace of acceleration!

For the jeep:
Acceleration of the jeep = (0.10 * 10^4 N) / (1400 kg)
Acceleration of the jeep = 0.07 m/s², a slightly speedier acceleration!

Lastly, let's find the horizontal push of the wheels of the jeep on the ground. Since there is no external force acting on the jeep horizontally, the horizontal push of the wheels on the ground is equal in magnitude to the net force applied by the tractor.

Horizontal push of the wheels of the jeep = |Net force|
Horizontal push of the wheels of the jeep = |-0.10 * 10^4 N|
Horizontal push of the wheels of the jeep = 0.10 * 10^4 N, the same magnitude but different direction as the force applied by the tractor!

So, there you have it! The instan-tee-giggles accelerations of the tractor and jeep, as well as the horizontal push of the wheels of the jeep on the ground. Keep the laughs rolling, my friend!

To calculate the instantaneous acceleration of an object, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration (F = m * a).

First, let's calculate the acceleration of the tractor. We know the mass of the tractor is 2000 kg, and the force applied is 1.50 * 10^4 N. Using Newton's second law, we can rearrange the formula to solve for acceleration: a = F / m.

Calculating the acceleration of the tractor:
a_tractor = (1.50 * 10^4 N) / (2000 kg)
a_tractor ≈ 7.5 m/s^2

The acceleration of the tractor is approximately 7.5 m/s^2.

Now, let's calculate the acceleration of the jeep. Both the tractor and the jeep are pulling on the rope, so they experience the same force. Since the mass of the jeep is 1400 kg, we can use the same formula to find its acceleration.

Calculating the acceleration of the jeep:
a_jeep = (1.50 * 10^4 N) / (1400 kg)
a_jeep ≈ 10.71 m/s^2

The acceleration of the jeep is approximately 10.71 m/s^2.

To calculate the horizontal push of the wheels of the jeep on the ground, we need to consider Newton's third law of motion. According to this law, for every action, there is an equal and opposite reaction. The force exerted by the wheels of the jeep on the ground is the reaction force to the force applied by the tractor's wheels.

Therefore, the horizontal push of the wheels of the jeep on the ground is equal to the force applied by the tractor's wheels, which is 1.60 * 10^4 N.

The horizontal push of the wheels of the jeep on the ground is approximately 1.60 * 10^4 N.