An object is pulled from rest by a constant, horizontal force of magnitude 6.25 N on a horizontal frictionless surface. In 12 seconds, the object gains a speed of \( 3.0 \mathrm{~m} / \mathrm{s} \). What is the mass of the object, in kg ?

To find the mass of the object, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \[ F = m \cdot a \] Where: - \( F \) is the force applied (6.25 N), - \( m \) is the mass of the object (in kg), - \( a \) is the acceleration of the object (in m/s²). First, we need to calculate the acceleration of the object. We know that the object starts from rest and reaches a speed of \( 3.0 \, \text{m/s} \) in \( 12 \, \text{s} \). We can use the formula for acceleration: \[ a = \frac{\Delta v}{\Delta t} \] Where: - \( \Delta v \) is the change in velocity (final velocity - initial velocity), - \( \Delta t \) is the time interval. Since the object starts from rest, the initial velocity is \( 0 \, \text{m/s} \), and the final velocity is \( 3.0 \, \text{m/s} \): \[ \Delta v = 3.0 \, \text{m/s} - 0 \, \text{m/s} = 3.0 \, \text{m/s} \] Now, substituting the values into the acceleration formula: \[ a = \frac{3.0 \, \text{m/s}}{12 \, \text{s}} = 0.25 \, \text{m/s}^2 \] Now that we have the acceleration, we can use Newton's second law to find the mass: \[ F = m \cdot a \] Rearranging for mass \( m \): \[ m = \frac{F}{a} \] Substituting the known values: \[ m = \frac{6.25 \, \text{N}}{0.25 \, \text{m/s}^2} = 25 \, \text{kg} \] Thus, the mass of the object is \( \boxed{25} \, \text{kg} \).