Calculating Normal Force, Acceleration, and Work in a Force and Work Problem

[logglypop]

A child pulls a 15kg sled containing a 5kg dog along a straight path on a horizontal surface. He exerts a force of a 55N on the sled at an angle of 20 degree above the horizontal. The coefficient of friction between the sled and the surface is 0.22 .
Find:a) normal force of the surface on the system , b) the acceleration of system, c) the work done by the child as the system move 7m.
This is how i approach them

a)normal force = Force of the child exert - Force of friction
55cos20 - .22( (15+5)9.8 ) = 8.563 N

b)normal force= ma
8.563N = ( 15+5 )a
a=.428m/s^2

c)W=Fd W=8.563(7)= 59.94J

give me suggest if I am make a mistake

 

[dynamicsolo]

A child pulls a 15kg sled containing a 5kg dog along a straight path on a horizontal surface. He exerts a force of a 55N on the sled at an angle of 20 degree above the horizontal. The coefficient of friction between the sled and the surface is 0.22 .
Find:a) normal force of the surface on the system , b) the acceleration of system, c) the work done by the child as the system move 7m.
This is how i approach them

a)normal force = Force of the child exert - Force of friction
55cos20 - .22( (15+5)9.8 ) = 8.563 N

You will need to think carefully about the vertical and horizontal components of the forces for this one. What you are calling the "normal force" here is the net horizontal force; the normal force is perpendicular to the horizontal surface.

While the force of kinetic friction is (mu_k)·(normal force), the normal force is not going to be equal in magnitude to the weight force of the dog and sled. What is the complete list of forces that contribute a vertical component?

 

[logglypop]

ah i see
the normal force on level surface is mg,so
a) F= 20(9.8) = 196N

B)
fk=ma
55cos20 - .22( (15+5)9.8 ) = 20a
a=.428 m/s^2

c)
W=Fd
55cos20 X 7= 361.78J

 

[dynamicsolo]

ah i see
the normal force on level surface is mg,so
a) F= 20(9.8) = 196N

The normal force would be equal to mg, if it weren't for the fact that there is another vertical force component acting on the sled. What is it?