Solve for Block Speed After 3m: F=50s^2, u=.3, v=2m/s

[joemama69]

Homework Statement

The force F, acting in a constant direction on the 20 kg block has a magnitude which varies with position s of the block. Determine the speed of the block after it slides 3 m. When s=0 the block is moving to the right at 2 m/s. The coefficient of kinetic friction is .3
F = 50s^2 and pushes on the crate from the hypot of a 3,4,5 triangle.

The Attempt at a Solution

I found the x,y components of force to be Fx = 40s^2, Fy = 30s^2
I found the Normal to be 30s^2 + 196.2 which i multiplied by u=.3 to get a frictional force in the oposite direction of 9s^2+58.86

I do not understand how to incorporate the initial velocity of 2 m/s, nore do i kow what equations to use. My book seems to hint at .5mv^2+Ps-uNs=.5mv^2 but i do not know what P is. Please help

 

[rl.bhat]

Find net Fx.
You can write f = dv/dt = dv/ds*ds/dt = v*dv/ds
So along x-axis v*dv = [40s^2 - ( 9s^2+58.86)]ds
Find the integration. To find the constant of integration c, use the condition , when s = 0 v= 2 m/s. Then to get the required answer put s = 3 m.

 

[joemama69]

Hey thanks a lot. I am pretty sure i got it but could u double check.

ok so i integrated and c=2, i pluged it in and solved for v

v=((80/3)s^3-6s^3-117.72s+2)^1/2 @ s=3, v=14.38 m/s

let me know how i did

 

[rl.bhat]

Your answer is correct.

 

[INTJ BSME]

Forgive me if I'm incorrect, but as I understand it that is the incorrect approach.

I believe the hint you described suggests that this problem should be solved using the work-energy relationship.

the first .5mv^2 refers to initial kinetic energy or ΣT(1) [this is where you use the initial velocity]
Ps-UNs refers to the change in work; P being the applied force (Fx), U being coefficient of friction, and s being displacement, or [ΣU(1-2)]
the last .5mv^2 refers to final kinetic energy, or ΣT(2)

I found a different final velocity using this method.
good luck!

 

[rl.bhat]

Hi INTJ BSME, welcome to PF.
Show your calculations.