How Much Force to Apply on a Lawnmower Handle at an Angle for Desired Velocity?

[danago]

Homework Statement

A lawnmower weighing 20kg has a handle attached at 35 degrees to the horizontal. If a man wishes to push the lawnmower, so that after 2.5s, its velcoity is 2 m/s, what force should he apply along the handle.

Homework Equations

<br /> \sum {\overrightarrow F } = m\overrightarrow a <br />

The Attempt at a Solution

ok, since after 2.5 seconds, it requires a velocity of 2 m/s, i found the acceleration of the mower to be 0.8 m/s/s. Since it has a mass of 20kg, the net force applied must then be:

<br /> \sum {\overrightarrow F } = 20 \times 0.8 = 16N<br />

So the horizontal component of the force he applies must be 16N in the direction of acceleration.

The horizontal component of the force (F) he applies down the handle will be F cos 35. So solving for F:

Fcos35=16

Gives F=19.5N.

According to the answer book, the answer is 120N. Where have i gone wrong?

Thanks,
Dan.

 

[Hootenanny]

If the question is quoted correctly, then I would agree with your answer. Does the text not mention any other factors, such as friction?

 

[danago]

Nope, no friction at all. What I've said is pretty much quoted word for word from the book.

Thanks for the reply.

 

[Ja4Coltrane]

yeah the book is def. wrong

 

[danago]

ok that's good to hear :) Thanks for the help