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

week 2 problem 7

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.

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