Math Problem Statement

A 65 kg sprinter, starting from rest, runs 45 m in 7.0 s at constant acceleration. What is the magnitude of the horizontal force acting on the sprinter?

Solution

To determine the magnitude of the horizontal force acting on the sprinter, we can break down the solution into two parts: finding the sprinter's acceleration and then using Newton's second law to calculate the force.

Step 1: Calculate Acceleration

Since the sprinter starts from rest, we can use the kinematic equation: d=12at2d = \frac{1}{2} a t^2 where:

  • d=45md = 45 \, \text{m} (distance traveled),
  • t=7.0st = 7.0 \, \text{s} (time taken),
  • aa is the acceleration.

Rearranging for aa: a=2dt2a = \frac{2d}{t^2} Substitute the values: a=2×45m(7.0s)2=90491.84m/s2a = \frac{2 \times 45 \, \text{m}}{(7.0 \, \text{s})^2} = \frac{90}{49} \approx 1.84 \, \text{m/s}^2

Step 2: Calculate Force

Using Newton's second law, F=maF = ma, where:

  • m=65kgm = 65 \, \text{kg} (mass of the sprinter),
  • a1.84m/s2a \approx 1.84 \, \text{m/s}^2.

Thus: F=65kg×1.84m/s2119.6NF = 65 \, \text{kg} \times 1.84 \, \text{m/s}^2 \approx 119.6 \, \text{N}

Answer

The magnitude of the horizontal force acting on the sprinter is approximately: 119.6N\boxed{119.6 \, \text{N}}

Would you like more details or have any questions?


  1. What if the mass of the sprinter changed to 70 kg?
  2. How would the force change if the acceleration were doubled?
  3. What if the sprinter covered 50 m instead of 45 m in the same time?
  4. How would air resistance affect the actual force needed?
  5. How is the force affected by running on a sloped surface?

Tip: Always check if the motion is under constant acceleration when using kinematic equations like d=12at2d = \frac{1}{2} a t^2.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Kinematics
Newton's Second Law of Motion
Constant Acceleration

Formulas

d = (1/2) * a * t^2
F = m * a

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 9-12