Question

The coefficients of static and kinetic friction between a 50 kilogram box and a horizontal surface are 0.60 and 0.40, respectively. (a) What is the acceleration of the object if a 250 newton horizontal force is applied to the box? (b) What if the applied force is 350 N?

Answer 1

Note that what matters here is uk = 0.40, as the block is moving.          
          
Thus, the sum of forces along y is          
          
Fy = N - mg = 0 ---> N = mg          
          
Thus,          
          
N = mg =    490   N  
          
Thus, the frictional force is          
          
Ff = uk N =    196   N  
          
Thus, the net force along x is          
          
Fx = Fapplied - Ff

= 250 N - 196 N =    54   N  
          
Thus, as Fx = m a, then a = Fx / m          
          
a =    1.08   m/s^2   [ANSWER, PART A]
          
FOR PART B:          
          
If the applied force is 350 N, then          
          
Fx = 350 N - Ff =    154   N  
          
Thus, the new acceleration is          
          
a =    3.08   m/s^2   [ANSWER FOR PART B]