Dynamics and Newton's Laws Help

[ArthurYan]

I have been absent for the past few days and he expects the class to these question but I was absent while he was teaching this so I have NO idea with to do here.

1st question: A 745 kg sled, initially at rest on a frictionless horizontal surface, is subjected to a force of 22500 N (E) for 10.5 s. The sled is then allowed to coast for an additional 14.5s. While the sled is coasting, determine the force required in order to stop it within a 500.0 m distance.

2nd question: A 2.00 kg ball is released (from rest) into a tank filled with water. The buoyant forces acting on the ball total 11.6 N (up). The ball takes 3.70 s to reach the bottom of the tank. Determine the net force acting on the ball, the ball's acceleration, and the depth of the tank.

3rd question: Catchy parachutes to Earth at a constant velocity of 15.4 m/s (down). Cathy has a mass of 57.2 kg and her parachute has a mass of 22.8 kg. What is the net force acting on Cathy, and determine the force the parachute applies on Cathy.

Last question: A 1250.0 kg car pulls a 250.0 kg trailer by applying a 3500.0 N force backwards on the road. The resistant forces (friction, wind resistance, etc.) on the trailer are 200.0 and on the car are 700.0 N. Determine the acceleration of the car and trailer together, and determine the force the car exerts on the trailer.

Thanks for any help guys!

 

[Nate810]

1st question: The force required to stop the sled within 500.0 m is 22500 N.2nd question: The net force acting on the ball is 8.6 N (down), the ball's acceleration is 2.33 m/s^2 (down), and the depth of the tank is 1.94 m.3rd question: The net force acting on Cathy is 16.4 N (down) and the force the parachute applies on Cathy is 16.4 N (up).Last question: The acceleration of the car and trailer together is 0.90 m/s^2 (backwards) and the force the car exerts on the trailer is 3500.0 N (backwards).

 

[ogb p]

Hello there,

I can understand your confusion with these questions since you were absent during the teaching of these concepts. Let me try to explain them to you.

1st question: This question is related to Newton's Second Law, which states that force is equal to mass multiplied by acceleration (F=ma). In this case, the sled has a mass of 745 kg and is initially at rest. It is then subjected to a force of 22500 N for 10.5 seconds. Using the formula F=ma, we can calculate the acceleration of the sled during this time. Once the force is removed and the sled starts to coast, we need to determine the force required to stop it within a distance of 500.0 m. This can be calculated using the formula F=ma, where "a" is the negative acceleration needed to bring the sled to a stop within the given distance.

2nd question: This question involves the concept of buoyancy, which is the upward force exerted by a fluid on an object immersed in it. In this case, the ball is released into the water and experiences a buoyant force of 11.6 N (up). We can use Newton's Second Law to determine the net force acting on the ball, and then use the formula F=ma to calculate the acceleration of the ball as it falls to the bottom of the tank. The depth of the tank can be calculated using the formula d=gt^2/2, where "d" is the depth, "g" is the acceleration due to gravity, and "t" is the time taken for the ball to reach the bottom.

3rd question: This question also involves the use of Newton's Second Law. We know that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, Cathy with a mass of 57.2 kg is falling at a constant velocity of 15.4 m/s. This means that the net force acting on her is zero, since there is no acceleration. However, her parachute with a mass of 22.8 kg is also falling at the same velocity. Therefore, the net force acting on Cathy and her parachute together is equal to the combined mass of both objects multiplied by their common velocity. The force applied by the parachute on Cathy can be calculated using the formula F=ma.

Last question: This question involves the concept of net force, which is the sum of all