• Support PF! Buy your school textbooks, materials and every day products via PF Here!

A 0.50 kg object is at rest. A 3.29 N force

  • Thread starter mustang
  • Start date
169
0
Problem 10.
A 0.50 kg object is at rest. A 3.29 N force to the right acts on the object during a time interval of 1.60 s.
At the end of this interval, a constant force of 4.07 N to the left is applied for 2.75 s.
b. What is the velocity at the end of the 2.75 s? Answer in m/s.
Note: Is the answer -22.385 m/s since in the question it says the force to the left?
 
Last edited by a moderator:

HallsofIvy

Science Advisor
Homework Helper
41,711
876
Okay, what have YOU done on this problem so far? What formulas do you think might apply?
 
169
0
I believe that the impulse-momentum theorem can be used here to slove the problem.
 

HallsofIvy

Science Advisor
Homework Helper
41,711
876
F= ma. You know the force and you know the mass so you can calculate the acceleration. Of course, the change in velocity is "acceleration * time".

To answer your original question, no, the "answer" is not -22.385 m/s. -22.385 m/s is the change in speed due to the second force. What was the speed when the second force was applied?
 
169
0
Would you add 2.75s to 1.60s to get 4.35s? From that would you then divide 0.50kg from 4.07N, to get 8.14 which is mulitplied by 4.35 to get the answer 35.409 m/s.
 

Related Threads for: A 0.50 kg object is at rest. A 3.29 N force

Replies
3
Views
1K
Replies
10
Views
4K
Replies
2
Views
1K
  • Posted
Replies
16
Views
6K
Replies
3
Views
3K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top