Finding Time with Work and Distance

[vuhwhat]

Homework Statement

A person pulls on a 20kg lawnmower with a force of 25.0N and moves the mower 4.50metes. The handle angle is 15* with the horizontal and μ=0.100.

What I've found:

a) Work done by person = 109J
B) Work done by friction= -85.3J
C) Find the time= ?

Relevant Equations

P=W/T
W= Fd

I tried using the 109 J to convert to seconds but that didn't work. I also looked through the energy equations but nothing worked. I am not sure how to find time.

 

[TSny]

Can you relate the total work done to the final velocity of the mower?

The time is related to the distance traveled and the average velocity.

 

[kuruman]

Or you can find the acceleration from the net horizontal force and use the kinematic equation relating acceleration, distance and time. It is appropriate to assume that the lawnmower starts from rest.

 

[vuhwhat]

Can you relate the total work done to the final velocity of the mower?

The time is related to the distance traveled and the average velocity.

F=ma --> a=F/m...
a= 25N/20kg
a= 1.25 m/s^2..

Acceleration is the change in speed so a=vt.. But I'm not sure how I find velocity?

 

[Delta2]

I think you got to take into account the force of friction as well, in order to calculate the net acceleration.

Once you find the net acceleration $a$, use the equation $s=v_0t+\frac{1}{2}at^2$ to find the time. We assume that the lawnmower starts from rest, that is $v_0=0$, I am not sure if this is a perfectly safe assumption though.

 

[haruspex]

you got to take into account the force of friction

and the angle at which the tractive force is applied. And don't forget to calculate the normal force correctly.

 

[Delta2]

and the angle at which the tractive force is applied. And don't forget to calculate the normal force correctly.

Oh damn how could I forget all this...Yes especially the normal force which won't be equal to the weight...

 

[jbriggs444]

The problem setter apparently intends that the tractive force is applied along the axis of the handle, though most mowers are equipped with stops that prevent handle travel beyond a certain range and thus permit forces with other directions as well.

 

[kuruman]

OP has correctly calculated the works in parts (a) and (b) which cannot be done without the correct expression of the normal force. The connection that OP missed is that, in this case,
$$W_{\text{net}}=W_{\text{F}}+W_{\text{fr.}}=F_{\text{net,x}}~\Delta x=m~a~\Delta x~\implies~a =\frac{W_{\text{F}}+W_{\text{fr.}}}{m\Delta x}.$$