Are we done with the problem or is there something else you don't understand?

]]>sorry I didn't realize shivams had posted something. I go to a college in Hampshire, not teeside.

please could you point out the other forum or thread as I don't want to have to post the same thing if its already been answered with advice

thanks

simon

This is about the 10th time I am doing this question from other forums. Let me guess: you go to Teeside university.

a)

Work=force ×distance

W=80 ×5

W=400J

b) Use sigma F = ma. From this find the sum of forces. There are two forces acting: the 80N push and the resisting force of friction. So force of friction = pushing force - ma.

c) Calculate the mass's acceleration from the data that it moves 5 m in 0.92 seconds: from d = 1/2 at^2 and solve for a. Then calculate velocity as a function of time: v = at. Finally Kinetic Energy as a function of time is KE= 1/2mv^2 = 1/2m(at)^2

d) Kinetic Energy as a function of distance is v^2 = 2ad so Kinetic Energy = 1/2mv^2 = 1/2m(2ad)

]]>What do you mean by a 'regular' graph ? Do you mean a straight line ?

ShivamS has shown how to calculate the work done. KE and work done are both measured in joules, but, in this question, they are not the same because some of the work is used to overcome friction and so the KE of the mass is less. Which were you asked for ?

Bob

]]>the graph that the tutor showed me looked like a regular graph not a quadratic.

thank you for your help

all the best

simon

it's the KE against time not the mass, so the question does make sense. As the object speeds up it gains KE so you can plot that against time.

I like to 'assemble' all the known facts like this:

initial velocity = u = 0

final velocity = v = to be determined

time taken = t = 0.92 s

distance travelled = s = 5 m

acceleration = a = to be determined

effective force = applied force less force due to friction = to be determined.

normal reaction of surface = R = mg

Firstly to find v

So that answer is correct.

On a velocity / time graph, the velocity line is straight, going through the origin so

So

So you'll get a quadratic graph going through (0,0)

Now for the friction:

KE against distance ?

In this case u = 0 so

Hope that helps,

Bob

]]>c) Plot a graph of the kinetic energy of the mass against time. Explain your calculations and state formulae used.

Mass against time? Is mass changing with time? Why?

]]>I have these two questions

c) Plot a graph of the kinetic energy of the mass against time. Explain your calculations and state formulae used.

d) Plot a graph of the kinetic energy of the mass against distance. Explain your calculations and state formulae used.

I have used this formula for the first one,

Ek = ½ mv2 = ½ ma2t2

Ek is proportional to t2.

this for the second

Ek = ½ mv2 , but we can substitute, giving

Ek = mgy

in not too sure how to show the graphs, but is the above formula along the right lines?

also I have to explain the use of the formula. I think this is just explain why I used it and any transposing required. if anyone knows a better answer for this one please give some advice.

taken from here, http://colourpoint.co.uk/sample_files/physics_as.pdf

this is the first part of the question,

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

a) Calculate the total energy expended in the acceleration, I got 11.81 acceleration

b) Calculate the coefficient of kinetic friction between the mass and the surface. Suggest materials from which the block and table might be made in order to give such results. I got 0.155

thanks

simon