Question

A person is pulling a 40 kg block of ice across an icy surface at a 25 degree angle. If the person is pulling with a force of 42 N, what is the magnitude of the acceleration of the block of ice? you may ignore drag, but not friction. Us=.1 and Uk=.03

I got this wrong on the homework and my teacher said that the normal forces does not equal mg here and that the tension is "pulling up"

Answer 1

Given

The mass of the block is m = 40 kg
angle theta with which the block is being pulled is theta = 25 degrees

the applied force F = 42 N
the coefficient of frictions are Us=.1 and Uk=.03

As the force applied on the block then the force components are written as

   F = Fx i + Fy j

   F = F cos theta i + F sin theta j

   F = 42 cos 25 i + 42 sin 25 j

   F = 38.06 N i + 17.75 N j

the force of gravity is mg will be downward
Fy is along y axis and the normal force also along y axis (upward)
net forces along y axis are must be equal to zero ,if the block is moving along X direction

   F_N+Fy -mg = 0

   F_N = mg- Fy

   F_N = mg - F sin theta

so the normal force is N = mg - F sin theta

as the block is in motion so that we should consider the coefficient of kinetic friction only

writing the force equation
  

   Fx - f = ma
   ma = F cos theta - mue_k*F_N

   ma = F cos theta - mue_k(mg - F sin theta)

   a = (F cos theta - mue_k(mg - F sin theta))/m

substituting the values

   a = (42 cos 25 - 0.03(40*9.8-42sin25))/(40) m/s^2

   a = 0.6709 m/s^2

the acceleration of the block is a = 0.6709 m/s^2