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Mazyrski [523]
3 years ago
7

An object sits at the edge of a table top with a mass of 180 kg. It is connected by a magical physics thread to a mass of 250 kg

via a pulley which is hanging in the air below the edge of the table top. There is an unknown amount of friction between the 180 kg mass and the table. The entire thing is accelerating at 2 m/s/s towards the edge of the table (250 kg mass is moving down).
What is the force of friction?
What is the coefficient of friction?

Physics
1 answer:
ratelena [41]3 years ago
5 0

Answer:

Force of Friction = 1640 N

Coefficient of friction = 0.911

Explanation:

forces act on the objects as shown in the graph.

Fx=Foroce of friction

T=tension of the thread

R=Normal Force(R=1800N)

μ= friction coefficient

a=acceleration = 2ms^{-2}

applying F=ma to 250kg object downwards

F=m*a\\2500-T=250*2\\T=2000N

applying F=ma to 180kg object rightwards

F=m*a\\\\T-F_x=m*a\\2000-F_x=180*2\\F_x=1640N

According to Frictional Force Formula,

F_x=μ*R

1640=μ*1800

μ=0.911

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Answer:

a   W_{3.5} = 490 \  J

b  W_{2.5} =  250 \  J

Explanation:

Generally the force constant is mathematically represented as

       k  = \frac{F}{x}

substituting values given in the question

=>   k  = \frac{16}{0.2}

=>   k  =  80 \ N /m

Generally the workdone  in stretching the spring 3.5 m is mathematically represented as

       W_{3.5} =  \frac{1}{2}  *  k  *  (3.5)^2

=>     W_{3.5} =  \frac{1}{2}  *  80  *  (3.5)^2

=>    W_{3.5} = 490 \  J

Generally the workdone  in compressing the spring 2.5 m is mathematically represented as

        W_{2.5} =  \frac{1}{2}  *  k  *  (2.5)^2

=>      W_{2.5} =  \frac{1}{2}  *  80 *  (2.5)^2

=>       W_{2.5} =  250 \  J

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Answer:

option (D)

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If we use

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In this equation, time is not included, so it is not useful.

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