4.5 | In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of

In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

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[carsmiles]

One of the most underrated human being on the planet earth!

[rafinharc]

Thank you so much, once again!! I found out another classmate has been watching your videos too! You make so much easier! We are from Montreal, Canada. Keep going, sir!

[chulachose]

Awesome video thanks for posting this really helped a lot!

[IamCodking]

Are you a prof? If not you should be! Thanks for the help over and over again!

[Saltydogwilm]

In the solution guide published by my professor the answer came out to be 0.36m.
They solved for acceleration using the a=F/m which was 75N/24kg =3.1 m/s^2.
Then they used V^2=V0^2+2ax which they deduced to be x=0.36m.

Could you provide any clarity? Thank you so much for your videos you all are saints.

[jenromeave]

Sir, why did you use friction for the acceleration when net external force is there? Why use -24N and not 51N? Thank you for your big help to all of us students. 💙

[DennaTadayon]

you actually lost a sig fig in the end there becase 2.25 is all significant so its 3 final sig figs not 2