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ipn [44]
3 years ago
13

At what point on the position-time graph shown is the object's instantaneous velocity greatest?

Physics
2 answers:
ioda3 years ago
7 0

I'm probably going to have to say C. E as it seems the steepest right around there. If I'm wrong on that, it has to be B. B

gizmo_the_mogwai [7]3 years ago
7 0

Answer:

The velocity is greatest at B.

Explanation:

The velocity or instantaneous rate of change is represented on a sketch by the slope of the position curve at a point.

Analyzing the points.

The slope at points F or D have slope as 0, since we have flat tangent lines for them.

The slope for point E is negative, since the curve is decreasing.

The slope for point B is positive, since the curve is increasing at B.

Thus the greatest value of the velocity is point B due the curve is increasing at that point, and the velocity is represented by its slope.

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How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

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scZoUnD [109]

Answer:

attached below

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