A student moves a box of books down the hall by pulling a rope attached to the box…?

A student moves a box of books down the hall by pulling a rope attached to the box. The

First, the forces

The Weight of the Box is

35[Kg]*9.8 [m/s^2] = 343 [N]

The Normal force is equal to the weight minus the vertical component of the applied force

N = 343 [N]- 185*sin(25°) [N]
N = 264.816 [N] (I will use approximations)

The friction force is the normal force times the friction coefficient

F= 0.27*264.816 [N]
F= 71.5 [N]

And now the acceleration would result from the horizontal component of the applied force minus the friction force

T = (185*cos(25)-71.5) [N]
T = 96.167 [N]

So, given the second Newton’s Law

a=T/m so

a= 96.167[N]/35[Kg] = 2.748 [m/s^2]

Cheers!

Student Moves