AP Physics 1 Exam Question:

A block of mass 2 kg is placed on a horizontal surface. A force of 10 N is applied to the block at an angle of 30° with the horizontal surface. Consider the tension force and the normal force acting on the block.

1. Determine the tension force acting on the block.
2. Determine the normal force acting on the block.

Given: Mass of block (m) = 2 kg Force applied (F) = 10 N Angle with horizontal surface (θ) = 30°

Assumptions and simplifications:

• The block is in static equilibrium and not moving.
• The horizontal surface is frictionless.

Answer with Step-by-Step Explanation:

1. To determine the tension force acting on the block, we need to resolve the applied force into its horizontal and vertical components. The horizontal component of the force will contribute to the tension force.

   The horizontal component of the applied force is given by: F_horizontal = F * cos(θ) = 10 N * cos(30°) = 10 N * 0.866 ≈ 8.66 N

   Therefore, the tension force acting on the block is approximately 8.66 N.

2. To determine the normal force acting on the block, we need to consider the vertical component of the applied force and the downward gravitational force acting on the block.

   The vertical component of the applied force is given by: F_vertical = F * sin(θ) = 10 N * sin(30°) = 10 N * 0.5 = 5 N

   The gravitational force acting on the block is given by: F_gravity = m * g = 2 kg * 9.8 m/s² = 19.6 N

   Since the block is in static equilibrium, the normal force must balance the sum of the vertical component of the applied force and the gravitational force.

   Total vertical force = F_vertical + F_gravity = 5 N + 19.6 N = 24.6 N

   Therefore, the normal force acting on the block is 24.6 N.

The tension force acting on the block is approximately 8.66 N, and the normal force acting on the block is 24.6 N.