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A 40 kg box is dragged across a frictionless horizontal surface by a rope. The tension in the rope is 65 N and it is at an angle of 42° to the horizontal. What is the box's acceleration?

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Final answer:

The acceleration of a 40 kg box dragged across a frictionless horizontal surface by a rope at a 42° angle with a tension of 65 N can be found by resolving the tension into components and applying Newton's second law.

Step-by-step explanation:

The question involves determining the acceleration of a box being pulled on a frictionless surface, which is a classic physics problem involving Newton's second law of motion.

Step-by-step solution:

  1. First, resolve the tension into horizontal and vertical components since the force is at an angle to the horizontal. The horizontal component (Tx) is T*cos(42°), and the vertical component (Ty) will be T*sin(42°), where T is the tension in the rope.
  2. Since the surface is frictionless, only the horizontal component will affect the horizontal motion of the box. Therefore, we use Tx = T*cos(42°) to calculate the force in the horizontal direction.
  3. Apply Newton's second law, F = m*a, to find the acceleration (a). Rearrange to a = F/m and substitute in the horizontal component of the force for F.
  4. Calculate the numerical value of acceleration using the given values: a = (65 N * cos(42°)) / 40 kg.

By following these steps, you can find the box's acceleration on the frictionless surface.

User Jan Kreischer
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