At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
We have that for the Question " constant applied force Fp of 11.0 N pushes a box with a mass=7 kg a distance x=15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is .110." it can be said that
the final velocity vf is
[tex]vf=3.83m/s[/tex]
and the applied force Fnew is
[tex]F_{new}=3.43N[/tex]
From the question we are told
A constant applied force Fp of 11.0 N pushes a box with a mass=7 kg a distance x=15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is .110. Assuming the box starts from the rest, what is the final velocity vf of the box at 15.0 m point? If there were no friction between the box and the floor, what applied force F new would give the box the same final velocity?
Generally the equation for the Force is mathematically given as
F_p-\mu mg=ma
Therefore
11-0.110*7*9.8=7*a
a=0.49m/s^2
Using Newtons equation
v^2=u^2+2as
v^2=0+2*0.49*15
v=3.83m/s
Therefore
F=ma
F=0.49*7
F_{new}=3.43N
Therefore
the final velocity vf is
[tex]vf=3.83m/s[/tex]
and the applied force Fnew is
[tex]F_{new}=3.43N[/tex]
For more information on this visit
https://brainly.com/question/23379286
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.