a student pulls a 10 kg wagon with a constant 140 N force at an angle 45 degrees above the horizontal for a distance of 3.0 m. how much work did the student do on the wagon?
what type of motion is the wagon experiencing ( e.g, constant speed, acceleration, etc)?
To calculate the work done by the student on the wagon, we can use the equation:
Work (W) = Force (F) × Distance (d) × Cosine (θ)
Where:
- Force (F) is the magnitude of the force applied by the student on the wagon, which is 140 N in this case.
- Distance (d) is the distance over which the force is applied, which is 3.0 m.
- Cosine (θ) is the cosine of the angle between the force and the direction of motion, which is 45 degrees in this case.
Using this equation, we can plug in the values and calculate the work:
W = 140 N × 3.0 m × Cos(45°)
Now, let's calculate the work.
W = 140 N × 3.0 m × (Cos(45°))
Cos(45°) = √2 / 2 ≈ 0.707
W = 140 N × 3.0 m × 0.707
W = 296.82 Joules
Therefore, the student did approximately 296.82 Joules of work on the wagon.
Now, let's determine the type of motion the wagon is experiencing. Since the force applied by the student is at an angle above the horizontal, we can break it down into horizontal and vertical components. The vertical component of force does not contribute to the motion in the horizontal direction.
The horizontal component of the applied force is responsible for the acceleration or deceleration of the wagon. In this case, since the force is constant and applied in the direction of motion, the wagon experiences a constant speed motion. The student's force counteracts any frictional or resistive forces, allowing the wagon to maintain a consistent speed.