What distance will it travel before coming to rest?

An 8 kg object moves with an initial velocity of 4 m/s comes to a rest due to friction afer it travels a horizontal distance of 10 m. If the initial speed of the object is doubled, what distance will it travel before coming to rest? Use g= 10 m/s2.What distance will it travel before coming to rest?Use the work-energy theorem. That theorem says that the change in an object's KE equals the sum of the work done on the object by all forces.



The change in KE is (by definition):

final KE 鈭?initial KE

= 0 鈭?陆m(v_initial)虏



The work done by friction is:

鈭抎istance 脳 F_friction (negative because the direction of the force opposes the direction of motion).



Thus, by the work-energy theorem:

鈭捖絤(v_initial)虏 = 鈭抎istance 脳 F_friction



Or:

distance = 陆m(v_initial)虏/F_friction



Now, the values of "m" and "F_friction" are the same in both trials. The only variable that changes is the speed. So you can see from the form of the equation what happens to "distance" when the value of "v_initial" doubles. (OK, hint: when v_initial doubles, "distance" increases by a factor of 2虏.)What distance will it travel before coming to rest?Ok let's apply the formula of Galilei to determine the acceleration of our object.

Galilei formula: v^2 = Vi^2 +2ad. Vi is the initial velocity and v is the final velocity. In our case v=0 because the object stops. Vi=4 m/s.

So 0=16 +20a (d=10 m) =%26gt; a=-16/20=-4/5= - 0,8 m/s^2.

m*a = - Ff. Ff is the friction force and is the only force actioning. Ff = umg. =%26gt; a=-ug =%26gt; u=a/g=%26gt; u=0,08.

Now the second case. Again Galilei 0=64+2ad.

a is the same because u does not modifies. The object and the surface are the same.

So d=-64/-2*0,8 =%26gt;d=40 m