# Grantham PHY220 Week 2 Assignment Problem 7

### A 7.93 kg box is pulled along a horizontal surface by a force $F_p$ of 84.0 N applied at a $47^{\circ}$ angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box?

SOLUTION:

The free-body diagram of the box

Solve for the normal force

$\sum F_y=ma_y$

$F_N-mg+F_psin\left(47.0^{\circ} \right)=0$

$F_N-7.93\left(9.80\right)+84.0\:sin\left(47^{\circ} \right)=0$

$F_N=16.28\:N$

Solve for the friction force

$F_{fr}=\mu _kF_N=0.35\left(16.28\:N\right)=5.70\:N$

Solve for the acceleration in the horizontal direction

$\sum F_x=ma_x$

$F_p\:cos\:\left(47^{\circ} \right)-F_{fr}=7.93\left(a_x\right)$

$84.0\:N\cdot cos\:\left(47^{\circ} \right)-5.70\:N=7.93\:kg\:\cdot \left(a_x\right)$

$a_x=\frac{51.59\:N}{7.93\:kg}=6.51\:m/s^2$

Therefore, the acceleration of the box is 6.51 m/s2 along the horizontal surface.