Question

5. Kelvin applied a rightward force of 302 N to a 29 kg desk to accelerate it across the floor. The

Answer 1

The force of friction is 213.2 N

Explanation:

The frictional force acting on an object sliding on a surface is given by:

$F_f = \mu mg$

where:

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration of gravity

In this problem we have

$\mu=0.750$

m = 29 kg

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f = (0.750)(29)(9.8)=213.2 N$

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