Math Problem Statement

The force exerted on a 8.0-kg block is shown in the figure as a function of time. How far does the block travel in the first 10.4 s? Assume that the motion is one dimensional and that the velocity of the block at time t= 0 s is 0 m/s.

The force exerted on a 8.0-kg block is shown in the figure as a function of time. How far does the block travel in the first 10.4 s? Assume that the motion is one dimensional and that the velocity of the block at time t= 0 s is 0 m/s.

Solution

To calculate the distance traveled by the 8.0-kg block in the first 10.4 seconds based on the force-time graph, we can follow these steps:

  1. Find the acceleration as a function of time:
    Use Newton's second law, F=maF = ma, where FF is the force, mm is the mass, and aa is the acceleration. For this block: a=Fm=F8.0a = \frac{F}{m} = \frac{F}{8.0}

  2. Find the velocity as a function of time:
    The velocity is the integral of acceleration over time: v(t)=v0+0ta(t)dtv(t) = v_0 + \int_0^t a(t) \, dt Since the initial velocity is v0=0m/sv_0 = 0 \, \text{m/s}, the velocity at any time can be found by integrating the acceleration.

  3. Find the displacement:
    The displacement is the integral of velocity over time: x(t)=0tv(t)dtx(t) = \int_0^t v(t) \, dt

Step-by-step process:

We will break down the time intervals based on the force changes in the graph:

  • From 0 to 2 seconds: Force = 4 N
  • From 2 to 4 seconds: Force = 8 N
  • From 4 to 6 seconds: Force = 6 N
  • From 6 to 9 seconds: Force = 4 N
  • From 9 to 10.4 seconds: Force = 6 N

For each interval, we will calculate the acceleration, velocity, and displacement.

Step 1: Calculate accelerations for each interval.

  • From 0t<20 \leq t < 2 s: a=48=0.5m/s2a = \frac{4}{8} = 0.5 \, \text{m/s}^2
  • From 2t<42 \leq t < 4 s: a=88=1.0m/s2a = \frac{8}{8} = 1.0 \, \text{m/s}^2
  • From 4t<64 \leq t < 6 s: a=68=0.75m/s2a = \frac{6}{8} = 0.75 \, \text{m/s}^2
  • From 6t<96 \leq t < 9 s: a=48=0.5m/s2a = \frac{4}{8} = 0.5 \, \text{m/s}^2
  • From 9t<10.49 \leq t < 10.4 s: a=68=0.75m/s2a = \frac{6}{8} = 0.75 \, \text{m/s}^2

Let me now calculate the total displacement by integrating velocity over these intervals.The block travels a distance of approximately 37.39 meters in the first 10.4 seconds, and its velocity at the end of this time period is about 7.05 m/s.

Would you like further details on the calculations or the kinematic process?

Here are some related questions:

  1. How would the result change if the initial velocity was not zero?
  2. What would be the distance traveled if the force was constant?
  3. How can the work done on the block be calculated from the force-time graph?
  4. What is the average acceleration over the entire time period?
  5. How would friction affect the block's motion in this scenario?

Tip: Breaking complex problems into smaller intervals, like we did with time, often simplifies integration in physics problems.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Newton's Second Law
Integration
Piecewise Function

Formulas

F = ma
v(t) = v0 + ∫ a(t) dt
x(t) = ∫ v(t) dt

Theorems

Newton's Second Law of Motion

Suitable Grade Level

High School (Grades 9-12)