A force of 9.0N is applied to a body at rest on a smooth horizontal surface. if the body has a mass of 18.0Kg and it moves in the direction of the force, find the workdone in (a) the first four seconds. (b) the fourth second

Explanation:

In this case a body of 18 kg is placed on the smooth surface .

The distance covered covered by the body in four seconds can be found by the relation  S = u t + 1/2 a t²  

Here S is the displacement , u is the initial velocity and a is the acceleration of the body .  t is the time taken

In this case initial velocity u = 0 , because the body is at rest .

the acceleration is  force/mass = 9/18 = 0.5 m/sec²

The values are substituted in the above equation

Thus S = 0 + 1/2 x 0.5 x ( 4 )²  = 4 m

Therefore the body will move 4 m in four seconds

The work done = Force x displacement

= 9 x 4 = 36 J

The displacement in fourth second can be found by relation

S₄ = u + a/2 ( 2 n - 1 ) ; where n is the number of seconds , which is 4 in this case

Thus S₄ = 0 + 0.5/2 ( 2 x 4 - 1 )  = 3.5/2 = 1.75 m

Therefore the work done = 9 x 1.75 =   15.75 J

none, its a machine

15.3 cos 65 degrees trig