Question

a box is at rest on a floor. then a student drags the box across the floor. the diagram shows the horizontal forces acting on the box as it is being dragged.

50 n ← 10 kg → 58 n

based on newton’s second law, what is the velocity of the box 1 s after the student starts dragging it?

a) 0.2 m/s
b) 5.8 m/s
c) 0.8 m/s
d) 10.8 m/s

Explanation:

Step1: Calculate net force

The net force \( F_{net} \) is the difference between the applied force and the frictional force. So \( F_{net} = 58\ N - 50\ N = 8\ N \).

Step2: Find acceleration using Newton's second law

Newton's second law is \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration. Rearranging for \( a \), we get \( a=\frac{F_{net}}{m} \). Substituting \( F_{net} = 8\ N \) and \( m = 10\ kg \), we have \( a=\frac{8\ N}{10\ kg}=0.8\ m/s^2 \).

Step3: Calculate velocity after 1 second

Using the kinematic equation \( v = u+at \), where the initial velocity \( u = 0\ m/s \) (box is at rest initially), \( a = 0.8\ m/s^2 \), and \( t = 1\ s \). Substituting these values, we get \( v=0 + 0.8\ m/s^2\times1\ s = 0.8\ m/s \).

Answer:

C. 0.8 m/s