First slide

Friction on horizontal surface

Question

A block of mass m = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40 N is applied, the acceleration of the block will be (g = 10 $rn / s^{2}$ )

Moderate

A

$5.73 m / \sec^{2}$
B

$8.0 m / \sec^{2}$
C

$3.17 m / \sec^{2}$
D

$10.0 m / \sec^{2}$

Solution

Kinetic friction = $μ_{k} R = 0.2 (mg - F \sin 30^{0})$
= $0.2 (5 \times 10 - 40 \times \frac{1}{2}) = 0.2 (50 - 20) = 6 N$
Acceleration of the block
$= \frac{F \cos 30^{0} - Kinetic friction}{Mass}$
= $\frac{40 \times \frac{\sqrt{3}}{2} - 6}{5} = 5.73 m / s^{2}$

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