What Is the Stopping Distance of a Mobile Machine Pushed at 1.2 m/s?

[Johny Prime]

Homework Statement

Having trouble with the concept here and have gotten stuck!

A man is pushing a mobile machine down a corridor at a steady speed of 1.2 m/s. The machine’s mass is 55 kg. Another person who has been walking in front of the machine has suddenly stopped. If the man is able to apply sufficient force to stop the machine in 1.0 s., will the man avoid a collision if the person is 1.0 m ahead of the machine? (Assume that the man is applying the same magnitude of force throughout that time.)

Velocity (initial) = 1.2 m/s
Mass = 55 kg
Velocity is at 0 after 1 second
Distance between is 1 meter

Homework Equations

• Acceleration = (Vf-Vi)/time
• Force = acceleration X mass
• Velocity = Distance/time

The Attempt at a Solution

Vf= 0 (stopped)
Vi= 1.2 m/s
So now finding the acceleration (deceleration): (0-1.2m/s)/1s = -1.2 m/s²
Can now find the force required to stop: Force = -1.2 X 55 kg
Force = -66 Newtons

This is where I am stuck. I am not sure if I am supposed to use : Force X Distance = 1/2 mv² formula (kinetic energy and work etc) or I am missing something obvious? If my work is correct so far what should be my next step?

 

[Borg]

It's probably a lot easier to step back and think about it differently. If he stops in one second, what do you think the average speed would be during that time?

 

[Let'sthink]

I think the problem must mention the coefficient of kinetic friction. Assume that it is given and one can find the limiting force, say of magnitude, f which man is applying to move the machine at a constant speed of whatever value. Now man wants to stop the machine and s applies the same force now in opposite direction helping friction. So the force applied on the machine will be 2f = 66 as correctly calculated by you. So f > 33 N.

 

[Johny Prime]

It's probably a lot easier to step back and think about it differently. If he stops in one second, what do you think the average speed would be during that time?

The average velocity would be -1.2 m/s, wouldn't it? I'm still not seeing the connection between this and finding the distance.

 

[gneill]

You're given that the machine stops in 1.0 seconds going from 1.2 m/s to 0 m/s in that time. If the man applies a constant force to achieve this then the acceleration will be constant and all the SUVAT formulas are available to use. Or just draw a velocity vs time graph and reminisce about that lesson where the prof talked about the area under a curve...

 

[Johny Prime]

It's probably a lot easier to step back and think about it differently. If he stops in one second, what do you think the average speed would be during that time?

Wait, I think I've figured it out. The average velocity would be 0.6 m/s right? Meaning that he travels 0.6 m? Is that correct?

 

[gneill]

Wait, I think I've figured it out. The average velocity would be 0.6 m/s right? Meaning that he travels 0.6 m? Is that correct?

Yes.

 

[Johny Prime]

Yes.

Thank you so much for your help, I think I understand it now.

 

[gneill]

Thank you so much for your help, I think I understand it now.

You're welcome.

 

[Let'sthink]

Did down mean down the slope. Then it is ok I can understand, otherwise just suvat based questions without description of the situation is not good for developing correct concepts.

 

[rodgerr]

I don't understand how you worked out the average velocity to be 0.6 m/s. Could you please explain this?

 

[CWatters]

I don't understand how you worked out the average velocity to be 0.6 m/s. Could you please explain this?

Hint:
Draw a graph of velocity vs time.
At t=0 v=1.2m/s
At t=1 v=0m/s
At constant acceleration the slope is constant/straight line between these two points.

 

[CWatters]

You can also solve this problem using the SUVAT equation to calculate the stopping distance..

S = (U+V)t/2
= (1.2+0)*0.5
= 0.6m

The (U+V)/2 part is the average velocity.