# Chapter 2 - Section 2.7 - Contact Forces - Practice Problem - Page 54: 2.14

(a) The magnitude of the frictional force is $111~N$ (b) The weight of the mat is $230~N$

#### Work Step by Step

(a) Since the mat is moving with a constant velocity, the force of kinetic friction opposing the motion must be equal in magnitude to the horizontal component of the groundskeeper's force. We can find the magnitude of the frictional force: $F_f = (120~N)~cos~22^{\circ} = 111~N$ The magnitude of the frictional force is $111~N$ (b) We can use the frictional force to find the weight of the mat: $F_N~\mu_k = 111~N$ $[mg-(120~N)~sin~22^{\circ}]~\mu_k = 111~N$ $mg-(120~N)~sin~22^{\circ} = \frac{111~N}{\mu_k}$ $mg = \frac{111~N}{\mu_k}+(120~N)~sin~22^{\circ}$ $mg = \frac{111~N}{0.60}+(120~N)~sin~22^{\circ}$ $mg = 230~N$ The weight of the mat is $230~N$.

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