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Leto [7]
2 years ago
11

A construction crew must use a lever to lift a boulder. The mechanical advantage of the lever is 6 and the lever applies a force

of 1600 N to the rock. What is the force that the construction crew must apply to the lever?
Physics
1 answer:
fredd [130]2 years ago
7 0

Answer:

The force the construction crew must apply to the lever is 266.67 N

Explanation:

Mechanical advantage is the measure of the ratio of output force to input force in a system. Hence, mechanical advantage (MA) is given by the formula

MA = \frac{F_{o} }{F_{i} }

Where MA is the mechanical advantage

F_{o} is the output force

and F_{i} is the input force

From the question, the mechanical advantage of the lever is 6

That is, MA = 6

and the lever applies a force of 1600 N to the rock

That is, Output force, F_{o} = 1600 N

To determine the force the construction crew must apply to the lever, that is the input force, F_{i}

From

MA = \frac{F_{o} }{F_{i} }

We can write that

F_{i}  = \frac{F_{o} }{MA}

Then,

F_{i} = \frac{1600N}{6}

F_{i} = 266.67 N

Hence, the force the construction crew must apply to the lever is 266.67 N.

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Check attachment for solution

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Read 2 more answers
A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe
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Answer:

(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

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         potential energy of the spring = 0.5 x k x l^{2}  (k= spring constant)

 therefore we now have

              mg(1.05 + l)  = 0.5 x k x l^{2}

              1.6 x 9.8 x (1.05 + l)  = 0.5 x 300 x l^{2}

               15.68 (1.05 + l) = 150 x l^{2}

                   16.5 + 15.68l = 150l^{2}

l = 0.39 m        

(B)   with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

     1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l)  = 0.5 x 300 x l^{2}

         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

            15.71 + 14.93l = 150^{2}

            l =0.38 m  

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      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

        2.608 (1.05 +l) = 0.5 x 300 x l^{2}

        2.74 + 2.608l = 150 x l^{2}

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