A child pulls a 15 kg sled containing a 5.0 kg dog along a straight path on a horizontal surface. He exerts a force of 55 N on the sled at an angle of 20 degrees above the horizontal. The coefficient of friction between the sled and the surface is 0.22.Calculate the word done by the child's pulling force as the system moves a distance of 7.0 m.

Respuesta :

Answer:

361.8 J

Explanation:

[tex]F[/tex] = Pulling force applied on the sled by the child = 55 N

[tex]\theta[/tex] = Angle between the direction of pulling force and displacement = 20 deg

[tex]d[/tex] = Displacement of the system = 7.0 m

[tex]W[/tex] = Work done by the child's pulling force

We know that work done is given as

[tex]W = F d Cos\theta\\W = (55) (7.0) Cos20\\W = 361.8 J[/tex]

The work done by child to move the system is 361.78 Joules.

The work done is given by an expression shown below,

                      [tex]Workdone=Force*Distance*Cos\theta[/tex]

Where [tex]\theta[/tex] is angle between force  and direction of displacement.

Given that, [tex]Force=55N,Distance=7m,\theta=20[/tex]

Substitute in above expression.

        [tex]Workdone=55*7*cos(20)\\ \\ Workdone=55*7*0.939\\ \\ Workdone=361.78Joule[/tex]

Thus, the work done by child to move the system is 361.78 Joules.

Learn more:

https://brainly.com/question/21854305