Mafer A.

asked • 02/16/21

Two forces are applied to a block

which undergoes a displacement of 8.50m to the right along a frictionless surface while the force acts. The applied force, B = 350.0N and pulls the block to the right. Work equal to 780.0J is done by these two forces. What is the force F?

θ = 30°

Osmany P.

Hello Mafer, could you add a picture of the problem please.
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02/16/21

1 Expert Answer

By:

Mafer A.

Hello, I apologize for not adding those details. F is 30°, in the negative x-axis.
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02/17/21

Dr Gulshan S.

tutor
Two forces are 350 N in +x and FSin30 = F/2 in -ve X Net force = 350 - F/2 Work done = Force * Distance = ( 350 - F/2) 8.5 = 780 F= 516.5 N Ask me if you have question Dr Gulshan Sawhney
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02/17/21

Mafer A.

I do not understand the process after you defined the equation for Work. At first, was that an inequality? Second, by F/2 you meant (F sin 30)? I've tried multiple times, but my calculations do not equal 516.5N. I do not follow the reasoning behind.
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02/17/21

Dr Gulshan S.

tutor
I can explain it to you with no first lesson fee
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02/17/21

Dr Gulshan S.

tutor
Please get back to me
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02/18/21

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