Question

A block with velocity v>0 slides along the floor (with no friction). It hits an ideal spring at time t=0 (configuration #1). The spring starts to compress until the block comes to a (momentary) stop (configuration #2). (Figure 1) (Later, the spring will of course expand, pushing the block back). Here we show you some plots relating to the motion of the block and spring. You will need to identify what these plots represent. In each plot, the point we label as "1" refers to configuration #1 (when the block first comes in contact with the spring). The point we label "2" refers to configuration #2 (which is the moment the block comes to rest, with the spring fully compressed). Here, "force" refers to the x-component of the force of the spring on the block and "position" (and "velocity") refer to the x-components of the position (and velocity) of the block. In all cases, consider the origin to be (0,0); that is, the x-axis represents y=0 and the y-axis represents x=0.

Part A

Look first a t graph A. (Figure 2)

Which of the choices given could this graph represent?

1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position

Part B

Now look at graph B. (Figure 3)

Which of the choices given could this graph represent?

1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position

Part C

Next look at graph C. (Figure 4)

Which of the choices given could this graph represent?

1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position

Answer 1

(A) position vs time

(B) Force vs position

(C) velocity vs time

Step-by-step explanation:

Part A

This graph shows that the position of the block increases with time along the x-axis exponentially (that is it increases in unequal amounts in equal time intervals). This is because the velocity of the block is changing with time and as a result the position changes in unequal amounts per time

PartB

The force on the spring increases in a negative direction going from zero to a negative value. This is because the spring is being compressed from configuration 1 to 2. The force of compression on a spring is usually taken to have a negative sign and expansion to have a positive sign. So in this case force becomes increasingly negative with time.

Part C

The velocity of the block decreases from a positive nonzero value (v>0) to zero because the spring resists the motion of the block. As a result the block comes to a stop momentarily. The velocity decreases exponentially because the acceleration of the block is also changing with time since the force of the block is decreasing with time.

Thank you for reading.