Work ang Energy problem (1 Viewer)

[Fat]

Work and Energy problem

Hi. Could someone please help me with a problem. It goes like this...
Solve Using the concepts of Work and Energy only (no Newtons laws, i.e forces and accelerations).
A 5.0kg mass, set sliding across a horizontal surface, then slides up a 30 degree slope. Initially the object is 4.0 m from the bottom of the slope and the slope itself is 6.00 m long. What initial velocity must the object have in order to reach the top of the slope:
a) if no frictioin
b) if the coefficient of sliding friction between the mass and surface is 0.2.

I have no idea how to start it, so any tips would be much appreciated.

 

[kini mini]

Do you understand gravitational potential energy? Find the difference between the GPE of the mass at the bottom and the top of the slope, the minimum kinetic energy of the mass must be equal to this.

Quickly googling to refresh my memory about the coefficient of friction (here), I don't remember this being in the HSC though i did it in science in year 10 . Work = Force x distance so you can use this to find the energy needed...dunno how you can deal with friction without forces though ?

 

[wogboy]

I don't recall the coefficient of friction ever being in the HSC/prelim Physics course, it's more of a uni style physics question, but anyway:

a) No friction

gravitational potential energy = kinetic energy

let gravtiational potential energy be W,

W = mgh
= 5 * 9.8 * 6 * sin30
= 147J

therefore, the kinetic energy K is,

K = 147J

but K = mv^2/2, so

v = sqrt(2K/m)
= sqrt(2 * 147 / 5) m/s
= sqrt(58.8) m/s
= 7.67 m/s

b) coefficient of friction, u = 0.2

therefore, work done by the friction force while on the flat surface is:

W1 = u * weight * distance across the flat surface
= 0.2 * (5 * 9.8) * 4
= 39.2 J

work done by the friction force whilst sliding up the slope is:

W2 = u * normal reaction force * distance up the slope
= 0.2 * (5 * 9.8 * cos30) * 6 J
= 58.8 * sqrt(3)/2 J
= 29.4 * sqrt(3) J

the gravitational potential energy of the mass at the top of the slope:

W3 = mgh
= 5 * 9.8 * 6 * sin30
= 147J

Now the required kinetic energy, K = W1 + W2 + W3, so

K = 39.2 + 29.4 * sqrt(3) + 147
= 186.2 + 29.4*sqrt(3)

but K = mv^2/2, so

v = sqrt(2K/m)
= sqrt(2 * (186.2 + 29.4*sqrt[3]) / 5) m/s
= 9.74 m/s