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DanielleElmas [232]
3 years ago
11

The principles used to solve this problem are similar to those in Multiple-Concept Example 17. A 205-kg log is pulled up a ramp

by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 25.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.855, and the log has an acceleration of 0.647 m/s2. Find the tension in the rope.
Physics
1 answer:
malfutka [58]3 years ago
8 0

Answer:2538.43 N

Explanation:

Given

Mass of Log=205 kg

Ramp angle=25^{\circ}

Coefficient of Friction=0.855

log acceleration=0.647 m/s^2

Let T be the Tension

T-mgsin\theta -f_r=ma

Here f_r=frictional\ Force=\mu N=0.855mgcos\theta =1556.76 N

T=mgsin25+m\times 0.647+1556.76

T=849.04+132.63+1556.76=2538.435 N

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Answer:

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Explanation:

Conceptual analysis

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(v_{f} )^{2} =(v_{i} )^{2} -2*g*y

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Known data

For v_{f} = 25\frac{ft}{s} ,y=\frac{1}{4} h; where h is the maximum height

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Problem development

We replace  v_{f} = 25\frac{ft}{s}  , y=\frac{1}{4} h (ft) in the formula (1),

[25^{2} =(v_{i} )^{2} -2*g*\frac{h}{4}   Equation (1)

in maximum height(h): v_{f} =0, Then we replace in formula (1):

0=(v_{i} )^{2} - 2*g*h

2*g*h=(v_{i} )^{2}

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We replace (h) of Equation(2) in the  Equation (1) :

25^{2} =(v_{i} )^{2} -2g\frac{\frac{(v_{i})^{2}  }{2g} }{4}

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