Question

While a roofer is working on a roof that slants at 36.0 ° above the horizontal, he accidentally nudges his 86 0 N toolbox, causing it to start sliding downward, starting from rest If it starts 4.00 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Answer 1

Answer:

Explanation:

mass of the toolbox = 86 / 9.8

= 8.775 kg

Net force acting on the toolbox down the roof

= mgsin36 - kinetic friction

= 86 x .5877 - 22

= 28.54 N

acceleration of box = 28.54 / 8.775

a = 3.25 m /s²

v² = u² + 2 a s

u = 0 , a = 3.25 , s = 4 m

v² = 2 x 3.25 x 4

v = 5.1 m / s

Answer 2

The velocity of the toolbox before it reaches the edge of the roof is 5.1 m/s.

The given parameters;

• angle of inclination of the roof, θ = 36⁰
• weight of the toolbox, W = 86 N
• length of the roof, s = 4 m
• kinetic friction force = 22 N

The mass of the toolbox is calculated from Newton's law;

W = mg

$m = \frac{W}{g} \\\\m = \frac{86}{9.8} \\\\m = 8.78 \ kg$

The net horizontal force of the toolbox is calculated as follows;

$\Sigma F_x = 0\\\\Wsin(\theta) - F_k = ma\\\\86sin(36) - 22= 8.78a\\\\28.57 = 8.78a\\\\a = \frac{28.57}{8.78} \\\\a = 3.25 \ m/s^2$

The velocity of the toolbox before it reaches the edge of the roof is calculated as;

$v^2 = u^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\v^2 = 2(3.25)(4)\\\\v^2 = 26\\\\v = \sqrt{26} \\\\v = 5.1 \ m/s$

Thus, the velocity of the toolbox before it reaches the edge of the roof is 5.1 m/s.

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