Please sort and match the following items.
A 2-kg toy car accelerates from 0 to 5 m/s^2. It travels 2 m.
A 2-kg toy car originally moving at 5 m/s is brought to rest. It travels 1m.
A 2 kg toy car is held at a height of 3.75 m above the floor.
A 2-kg toy car accelerates from 5 to 10 m/s^2. It travels 7 m.
A 2-kg toy car accelerates from 10 to 5 m/s^2. It travels 10 m.
No work is done on or by the object.
10 joules of work is done on the object.
20 joules of work is done by the object.
70 joules of work is done on the car.
100 joules of work is done by the car.
...accelerates from 0 to 5 m/s^2.....
HUH ??? maybe 5 m/s ?? Whatever . Please get it right.
That's how it was on the question. Idk.
Somebody need to find out.
To sort and match the given items, we need to analyze the information provided and find the corresponding answers.
First, let's consider the concept of work. Work is defined as the transfer of energy that occurs when a force is applied to an object and causes it to move in the direction of the force. The formula for work is:
Work = Force x Displacement x cos(theta)
where Force is the applied force, Displacement is the distance moved by the object, and theta is the angle between the applied force and the displacement.
Now, let's examine each statement and determine if work is done on or by the object:
1. A 2-kg toy car accelerates from 0 to 5 m/s^2. It travels 2 m.
No work is done on or by the object. The car doesn't move against any force, so neither work is done on the car nor by the car.
2. A 2-kg toy car originally moving at 5 m/s is brought to rest. It travels 1m.
20 joules of work is done by the object. When the car is brought to rest, it slows down, and the force of friction opposes the direction of motion. Therefore, work is done by the car to overcome the force of friction.
3. A 2 kg toy car is held at a height of 3.75 m above the floor.
70 joules of work is done on the car. When the car is lifted, work is done against the force of gravity. The formula for work done against gravity is:
Work = Force x Displacement x cos(theta)
In this case, the applied force is the weight of the car (mg), the displacement is the height of the lift, and the angle between the force and displacement is 0 degrees (cos(0) = 1). Therefore, the work done against gravity is equal to the weight of the car multiplied by the height: work = (2 kg) x (9.8 m/s^2) x (3.75 m) = 73.5 joules ≈ 70 joules.
4. A 2-kg toy car accelerates from 5 to 10 m/s^2. It travels 7 m.
No work is done on or by the object. Similar to the first statement, the car doesn't move against any force, so neither work is done on the car nor by the car.
5. A 2-kg toy car accelerates from 10 to 5 m/s^2. It travels 10 m.
100 joules of work is done by the car. In this case, the car is decelerating, and the force of friction acts in the direction opposite to the car's motion. Therefore, work is done by the car to overcome the force of friction.
To summarize the sorted and matched items:
No work is done on or by the object.
10 joules of work is done on the object.
20 joules of work is done by the object.
70 joules of work is done on the car.
100 joules of work is done by the car.