AP Physics 1 Exam Question:
A wooden block of mass 5 kg is placed on a frictionless horizontal surface. A string is attached to the block and pulled vertically upwards with a force of 30 N. The angle between the string and the horizontal surface is 30°. Calculate the tension in the string and the normal force acting on the block.
Answer:
To solve this problem, we need to consider the forces acting on the block.
Let's analyze the forces acting on the block:
Weight (mg): The weight of the block can be calculated using the formula W = mg, where m is the mass of the block and g is the acceleration due to gravity. In this case, the mass of the block is 5 kg, and the acceleration due to gravity is approximately 9.8 m/s². Therefore, the weight of the block is 5 kg × 9.8 m/s² = 49 N directed vertically downwards.
Tension (T): The tension in the string is the force applied to the block in the vertical direction. In this case, the applied force is 30 N, and the angle between the string and the horizontal surface is 30°. To find the vertical component of the applied force, we can use the formula T*cos(θ), where T is the magnitude of the tension force and θ is the angle between the force and the vertical direction. Therefore, the vertical component of tension is 30 N × cos(30°) ≈ 25.98 N directed upwards.
Normal Force (N): The normal force is the force exerted by a surface perpendicular to the surface of contact. Since the block is placed on a horizontal surface and there is no vertical acceleration (since it is frictionless), the normal force would be equal in magnitude and opposite in direction to the weight of the block. Therefore, the normal force is 49 N directed vertically upwards.
Hence, the tension in the string is approximately 25.98 N directed upwards, and the normal force acting on the block is 49 N directed vertically upwards.
Therefore, the tension in the string is 25.98 N directed upwards, and the normal force acting on the block is 49 N directed vertically upwards.