A horizontal force of 80N acts on a mass of 6 Kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant

Showing all working Calculate the total energy expanded in the acceleration;

S=(.5)at^2
5=(.5)a(0.92^2)
=0.4232 a
a=5/0.4232 = 11.8147 ms^2

V=at
=11.8147*0.92
=10.8695 ms^-1

s=(.5)a(t^2)
s=4.999 m

Work done;

W=F*d=80*5= 400J

Power;

W=P*t
P=W/t=400/0.92=434.78 watt

The final velocity of the mass
Vf^2=Vi^2+2ad (vi=0, as the mass is initially at rest)
F=m*a
a=F/m=80/6=13.33 m/s^2
Vf^2=2*13.33*5=133.33
Vf=11.55m/s

the final momentum
p=mv
=6*11.55=69.3 N/s

the final kinetic energy
K=0.5*m*v^2=533 J

Have I answered the fundamental points of the question of is there irrelevant things I could remove, I'm just trying to answer the question as thoroughly and best I can..thanks

"the final kinetic energy
K=0.5*m*v^2=533 J"

The final kinetic energy cannot be larger than the work done.

Secondly, you compute the acceleration as 10.8695 ms^-1 at the beginning of your analysis using data supplied in the problem. But near the end you compute it as Vf=11.55m/s using F=ma. What does this tell you?

I've re-done it..May be correct this time.
As The work done = 400 J and it says assume no energy is lost, and cannot exceed 400 J

W = F x d
= 80 x 5 = 400 J

Thus;

KE = 1/2*M*V2
400 = 1/2*6*V2
400 = 3*V2
400/3 = V^2
sqrt(400/3) = v
v = 11.547 m/s

a = v – u / t
a = 11.547 – 0 / 0.92
a = 12.551 m/s2

V = a*t

As the acceleration remains the same throughout and the velocity increases I used the equation V = a*t to change the velocity at various times from t=0 to t=0.92 to give me all my graph points (400J being the figure for t = 0.92 seconds for the equation KE = 1/2*M*V2)

on the right lines now? or have i gone wrong?

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question also states to Plot a graph of the kinetic energy of the mass against time thats why ive concentrated at individual points of time i.e t=0, t=0.92

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By using Newtons second law (a = 80/6) and s = 0.5 a t^2 distance travelled is more than that given in the problem.This implies the presence of sliding friction.Therefore work done or work expended must be calculated by using W = Fs = 80(5) = 400 J.If you use other data as you have done in the first part friction will also be included and hence you get net work done by all forces - not the work expended which implies the work done by external agent.

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Kinetic energy K = 0.5 m v^2 =0.5 m (at^2)^2 which is a parabola.

S=(.5)at^2
5=(.5)a(0.92^2)
=0.4232 a
a=5/0.4232 = 11.8147 ms^2

s=(.5)a(t^2)
s=4.999 m

The final velocity of the mass
Vf^2=Vi^2+2ad (vi=0, as the mass is initially at rest)
F=m*a
a=F/m=80/6=13.33 m/s^2
Vf^2=2*13.33*5=133.33
Vf=11.55m/s
Whats going on above?:surprised
Is the floor having some friction?

my first post was my first attempt whereby i went all over the place. is my second post right?

Please refer to my first post.
Attention , cupid.callin
Sliding friction must be present as explained in the first post.
Attention ,charger
you are equating work done by 80 N to Change in Kinetic energy.This is wrong because you are neglecting friction the presence of which is explained in my first post.

my first post was my first attempt whereby i went all over the place. is my second post right?
My advice: use acceleration you found by s = 0.5 at2 (Why ?)

ehild
Homework Helper
80 N is the force of some external agent. If you calculate the acceleration from time and displacement you will find that the net force is only 70.9 N. The difference is due to sliding friction. The work of the resultant force is equal to the change of KE.

ehild