Problem 1: A 2.0 kg box slides on a floor. A friction force of 5.0 N opposes the motion. If the box starts with a speed of 5.5 m/sec, how far does is slide before coming to rest? Answer: d = 6.1 m
Problem 1: A 2.0 kg box slides on a floor. A friction force of 5.0 N opposes the motion. If the box starts with a speed of 5.5 m/sec, how far does is slide before coming to rest? Answer: d = 6.1 m
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Kinetic Energy and Work Problems**
**Equation:**
- \( W_{\text{net}} = \frac{1}{2}m(v_f)^2 - \frac{1}{2}m(v_i)^2, \quad W_{\text{net}} = \sum_i W_i , \quad W = Fd \cos \theta \)
**Problem 1:**
A 2.0 kg box slides on a floor. A friction force of 5.0 N opposes the motion. If the box starts with a speed of 5.5 m/sec, how far does it slide before coming to rest?
*Answer:* \( d = 6.1 \, \text{m} \)
**Problem 2:**
What is the force a 60.0 kg sprinter exerts backward on the track to accelerate from 2.00 m/sec to 8.00 m/sec in a distance of 25.0 m, if they encounter a wind that exerts an average force of 30.0 N against them?
*Hint:* \( W_{\text{net}} = W_{\text{runner}} - W_{\text{wind}} \)
*Answer:* \( F_{\text{runner}} = 102.0 \, \text{N} \)
**Problem 3:**
A 500-kg dragster accelerates from rest to a final speed of 110 m/sec in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is the work done by the dragster?
*Hint:* \( W_{\text{net}} = W_{\text{dragster}} - W_{\text{friction}} \)
*Answer:* \( W_{\text{dragster}} = 3.5 \times 10^6 \, \text{J} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F916b3e43-3e70-4159-b0c4-23957999f35d%2Ffc51a647-d653-4b81-adb5-319bdef7a431%2Fychnuy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Kinetic Energy and Work Problems**
**Equation:**
- \( W_{\text{net}} = \frac{1}{2}m(v_f)^2 - \frac{1}{2}m(v_i)^2, \quad W_{\text{net}} = \sum_i W_i , \quad W = Fd \cos \theta \)
**Problem 1:**
A 2.0 kg box slides on a floor. A friction force of 5.0 N opposes the motion. If the box starts with a speed of 5.5 m/sec, how far does it slide before coming to rest?
*Answer:* \( d = 6.1 \, \text{m} \)
**Problem 2:**
What is the force a 60.0 kg sprinter exerts backward on the track to accelerate from 2.00 m/sec to 8.00 m/sec in a distance of 25.0 m, if they encounter a wind that exerts an average force of 30.0 N against them?
*Hint:* \( W_{\text{net}} = W_{\text{runner}} - W_{\text{wind}} \)
*Answer:* \( F_{\text{runner}} = 102.0 \, \text{N} \)
**Problem 3:**
A 500-kg dragster accelerates from rest to a final speed of 110 m/sec in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is the work done by the dragster?
*Hint:* \( W_{\text{net}} = W_{\text{dragster}} - W_{\text{friction}} \)
*Answer:* \( W_{\text{dragster}} = 3.5 \times 10^6 \, \text{J} \)
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