#629 A child pulls a 7.00 kg sled along a flat surface

A child pulls a 7.00 kg sled along a flat surface - Physics

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Free Chegg Question

A child pulls a 7.00 kg sled along a flat surface.  The coefficient friction between the sled and the snow is 0.100.  The string that the child pulls with is angles at 40 degrees.  If the sled's acceleration is 0.250 m/s2, with how much force is the child pulling?

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Best Free Chegg Answer

From the force diagram applying F=ma for the xand y components.

X comp: FtCosx – Fr = ma

      Or    Ft Cosx- µFn = ma -----------------(1)        

Ycomp:   Fn + FtSinx – mg =0

       Or       Fn= mg – FtSinx

Substituting in eq.(1) we get FtCosx - µ(mg –FtSinx) = ma

                 Or    Ft = [ m(a + µg)] /(Cosx + µSinx)}

                               =[7(0.250 + 0.1x9.8)]/(Cos40 +0.10 Sin40)

                          Ft = 10.4N

Free Chegg Answer 2

Frictional force = uN, where u is 0.1 and N is the weight of the body, which is 70N. Therefore, frictional force is 7N. Since the angle is 40 degrees, it's horizontal component , F cos 40, is equal to (ma friction) , which is 7.7 * 0.250 = 1.925 N. Hence, F = 1.925/cos 40. Which is 2.5 N

Free Chegg Answer 3

Frictional force = uN, where u is 0.1 and N is the weight of the body, which is 70N. Therefore, frictional force is 7N. suppose boy is pulling with force F Since the angle is 40 degrees, it's horizontal component , F cos 40, Now Fcos40 - 7N = m*a Fcos 40 = 7N + 7*.25 = 8.75N F = 8.75/ cos 40 = 8.75/.766 = 11.422 N