Box being pulled across horizontal floor with friction

AI Thread Summary
The discussion focuses on a physics problem involving a 30kg box being pulled on a horizontal surface with a force of 50N in the x-direction and 30N in the y-direction, while experiencing a frictional force of 20N. Using Newton's second law, the net force in the x-direction is calculated as 30N, leading to an acceleration of 10.89 m/s². The normal force acting on the box is determined to be 620.7N, accounting for both gravitational and vertical acceleration components. The coefficient of kinetic friction between the box and the floor is found to be 0.0322. The analysis emphasizes the importance of separating the x and y components of the applied force for accurate calculations.
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Homework Statement


Sam is pulling a box of mass 30kg on a horizontal floor with a force FT= <50N, 30N>. The frictional force acting on the box by the floor is 20N.
A) By using Newton's second law find the acceleration of the block.
B) What is the normal force acting on the block by the floor?
C) What is the coefficient of kinetic friction between the floor and the box?

Homework Equations


Fnet=Fx+Fy
F=ma
Fnormal=m(g+ay) (g is approximated gravity coefficient, 9.8m/s2)
μkFnormal=Fk(force of kinetic friction)
|F|=√Fy2+Fx2

The Attempt at a Solution



A)
FTX=50N - 20N = 30N
FTY=30N +294N = 324N
|F|=√(324N)2 + (30N)2 = 326.77N
F=ma
326.77N = (30kg)a
a = 10.89 m/s2

B)
Fnormal=m(g+ay)
Fnormal=(30kg)(9.8m/s2+10.89 m/s2)
= 620.7N

C)
μkFnormal=Fk
μk(620.7N) = 20N
μk = .0322
 
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You are given the x and y components of the applied pulling force. Leave it that way; do not try to combine the components into their resultant vector. Instead, since you know that the net force in the x direction is 30 N, you can now find the acceleration in the x direction using Newton 2. In the y direction, is there any acceleration in that direction if the box stays on the floor?
 
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