Answered

Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

If a box is pushed across a floor with a force of 130N. The frictional force acting between the box and the floor is 30N, over a time period of 5 sec,
Determine :
1) What is the net force acting on it?
2) Acceleration acting upon the box if it starts from rest and attains a certain Kinetic energy after being pushed 25m?
3) Also find the mass of the box.
4) What will be the kinetic energy and final velocity?

Sagot :

olemakpadu
The force acting in the front direction is the 130N.
The frictional force is acting backwards          30N.

1) The net force is 130N - 30N    =  100N

2)  s  = ut + (1/2)at^2             u = 0,  Start from rest,  s = 25m t =5.

25 = 0*5  +  (1/2)* a * 5^2.

25 = 0  +  25/2  * a.

25  =    (25/2)a.      Divide 25 from both sides.

1 =  (1/2)* a.          Cross multiply.

2 = a.

a = 2 m/s^2.

3) Mass of the box
Net Force,  F = ma
                   100 =  m*2.        Divide both sides by 2.
                    
                     100/2  =  m
                       50       =  m.
                        m  = 50 kg.

4)  Final velocity ,   v = u + at.
                                   v  =  0  + 2*5 = 10 m/s.
                                  
   Kinetic Energy,  K  =  (1/2) * mv^2.
                                    =    1/2  * 50 * 10 * 10.
                                    =      2500 J.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.