Question

2. Kevin works as a janitor, and he is pushing a fully-
loaded recycling bin that weighs 557 N across the
floor. He applies a force of 410 N at an angle of 15°
below the horizontal. The coefficient of kinetic friction
between the recycling bin and the floor is 0.46. How
long does it take him to move the bin 6.5 meters,
assuming the bin starts from rest?

Answer 1

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

• weight of the load, w = 557 N
• force applied , F = 410 N
• angle of force, =  15°
• coefficient of kinetic friction  = 0.46
• distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

• m is the mass of the recycling bin
• $F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

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