A 12 kg block is released from the top of an incline that is 5.0 m long and makes an angle of 40.0 degrees to the horizontal. A force of friction of 60.0 N impedes the motion of the box.
1. Find the acceleration of the box.
2. How long will it take the box to reach the bottom of the incline?
3. What is the coefficient of friction?

[tex]P_x -fr = ma\\a = \dfrac{P_x-fr}{m}\\a = \dfrac{mg\sin \theta -fr}{m}\\\\a = \dfrac{(12\;kg)(9.8\;m/s^2)\sin 40^\circ -60N}{12\;kg}\\\\a\approx1.3 \;m/s^2[/tex]

b) We know from cinematic that:

d = at²/2

Solving for t:

[tex]t = \sqrt{\dfrac{2d}{a}} =\sqrt{\dfrac{2(5\;m)}{1.3\;m/s^2}} = 2.8\;s[/tex]

c) We write the sum of forces on the y axis as:

N - Py = 0

N = Py

N = mgcosθ

fr = μN

Solving for μ:

μ = fr/N

[tex]\mu = \dfrac{fr}{mg\cos\theta} = \dfrac{60\;N}{(12\;kg)(9.8\;m/s^2)\cos40^\circ} =0.67[/tex]

andromache andromache · Answer 2 · 2022-03-18T15:20:37+01:00

1. The acceleration of the box is 1.3 m/s^2.

2. The time taken for reaching the bottom of the incline is 2.8 seconds.

3. The coefficient of friction is 0.67.

Calculation of the acceleration, time taken, and the coefficient of friction:

1.

We know that

Px - fr = ma

a = (px - fr)/m

a = (mg sin - fr)/m

= (12) (9.8) sin 0 - 60 N / 12 kg

= 1.3 m/s^2

2. The time taken should be

= √2d/a

= √295) / 1.3

= 2.8 seconds

c. The coefficient of friction should be

N - Py = 0

N = Py

μ = fr/mgcos

= 60N / (12kg) (9.8) (cos 40)

= 0.67

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