Figure Ex5.7 shows the velocity graph of a 5.0 kg box as it pushed to the right (->) along a horizontal surface. Assume that a 12 newton friction force acts on the box to the left (<-) during its the entire motion.
What is the amount of right-ward (->) acting push on this object at the given times?
t=2 N=?
t=7 N=?
To find the amount of right-ward (->) acting push on the object at the given times, we need to analyze the velocity graph and consider the forces acting on the object.
First, let's understand the information given:
- The object is a 5.0 kg box.
- The graph shows the velocity of the box as it is pushed to the right (->) along a horizontal surface.
- A friction force of 12 newtons acts on the box to the left (<-) during the entire motion.
From the graph, we can infer that the object is initially at rest, then its velocity increases steadily, and finally, it reaches a constant velocity.
At t = 2 s:
To find the magnitude of the right-ward acting push, we need to consider the net force acting on the box. At this point, the opposing force is the friction force acting to the left (12 N). Therefore, the right-ward acting push must be equal in magnitude but opposite in direction. So, the push at t=2 s is -12 N.
At t = 7 s:
Again, we need to consider the net force acting on the box. At this point, the friction force is still acting to the left (12 N). The box has reached a constant velocity, so its acceleration is zero. According to Newton's second law (F = m*a), when the acceleration is zero, the net force is also zero. Therefore, the right-ward acting push must be equal in magnitude but opposite in direction. So, the push at t=7 s is -12 N.
In summary, the right-ward acting push on the object at t=2 s and t=7 s is -12 N.