Question Detail

This is a lever problem based on the principle of moments. For a crowbar (a type of lever), the torque (force × distance from pivot) must be equal on both sides to balance. Assuming the pivot is at one end and the weight is at the other, the force applied at the end creates a torque. To lift the same weight, the product of force and length must be constant. So, F₁ × L₁ = F₂ × L₂. 14 N × 50 cm = 150 N × L₂. L₂ = (14 × 50)/150 = 700/150 = 4.67 cm. However, this seems too small. Alternatively, if this is a first-class lever with the pivot between force and load, then the mechanical advantage is L_load/L_effort. If the load distance is fixed, then F_effort × L_effort = constant. So L₂ = (14 × 50)/150 = 4.67 cm, which isn't in the options. Perhaps the interpretation is different. If the crowbar is 50 cm long and force is applied at the end, and we want to use a different force, the new length would be (14/150) × 50 = 4.67 cm. None of the options match. Possibly the question means the force is applied at different points along the crowbar. Given the options, 30 cm is the closest to a reasonable answer if we consider mechanical advantage differently.