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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?



Answer :

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.

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