Net Force on a Sled: 17.7 m/s^2 at 68.5°

[MCATPhys]

A 10 kg sled is pulled at 35 degrees east of north with 100 N. There is a force of 150 N due east. And a friction force of 50 N. What is the acceleration of the sled, and its direction?

So I basically added up all the forces in the x direction, and the y direction, and used the pythagorean theorem to get the net force, and divided it by the mass to get the acceleration. But somehow, I don't get the right answer, which is 17.3 m/s^2 at 68.5 degrees east of north.

My work:
Net x force = cos(55)(100) + 150 - 50 = 157.36 N
Net y force = sin(55)(100) = 81.9 N
tan (x) = 81.9/157.36; x = 27.5 degrees
square root of (157.36^2 + 81.9 ^2) = 177.4 N; 17.7 m/s^2

 

[rl.bhat]

You have to subtract the frictional force from the resultant force, because the frictional force acts in the opposite direction of the motion.

 

[MCATPhys]

You have to subtract the frictional force from the resultant force, because the frictional force acts in the opposite direction of the motion.

"Net x force = cos(55)(100) + 150 - 50 = 157.36 N"

But I did subtract it. I subtracted it from the x force, because it acts in the opposite direction of the x axis.

 

[MCATPhys]

Nevermind. I understood what you meant. I subtracted from the angled force and got the right answer... thank you so much!

 

[rdl]

tan (y) = 81.9/157.36; y = 27.5 degrees

I would first check my calculations and make sure I did not make any mistakes in my calculations. It is possible that I made a mistake in my trigonometric calculations or in adding up the forces. I would also double check the given values to ensure they are correct. If my calculations are correct, then I would consider the possibility of external factors such as air resistance or friction on the sled affecting the acceleration. I would also consider the possibility of the sled's mass being different from the given value of 10 kg. If all these factors are accounted for and the calculated acceleration still does not match the given answer, then it is possible that there is a mistake in the given answer. As a scientist, it is important to always double check and verify results to ensure accuracy.