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hichkok12 [17]
3 years ago
9

The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25

.0 m/s, use energy concepts to calculate the distance it will slide before coming to a stop. Show all your work starting with the most fundamental concepts/equations. That is, I have to know where your equations CAME FROM (and you know where they came from).
Physics
1 answer:
Vikki [24]3 years ago
6 0

Answer:

375 m.

Explanation:

From the question,

Work done by the frictional force = Kinetic energy of the object

F×d = 1/2m(v²-u²)..................... Equation 1

Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.

Note: If is negative because it tends to oppose the motion of the object.

Substitute into equation 2

d = 1/2(60)(0²-25²)/-50

d = 30(-625)/-50

d = -18750/-50

d = 375 m.

Hence the it will slide before coming to rest = 375 m

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a. v_f=1.477m/s

b. ΔK=1558.3J

c. E_k=1034.7 J

Explanation:

a).

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v_f=\frac{m_1*v_1+m_2*v_2}{m_1+m_2}

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d).

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The given parameters;

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