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A person pulls on a 7.4 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.6 m/s and speeded up to 2.6 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate?

Respuesta :

Answer:

Work done will be 59.54 J

Explanation:

We have given that person pulls a 7.4 kg crate against 22 Newton frictional force

So frictional force [tex]f=22N[/tex]

Balance force is given by [tex]F-f=ma[/tex]

We have given initial speed u = 1.6 m /sec

And final speed v = 2.6 m/sec

Distance s = 2 m

From third equation of motion we know that

[tex]v^2=u^2+2as[/tex]

[tex]2.6^2=1.6^2+2\times a\times 2[/tex]

[tex]a=1.05m/sec^2[/tex]

So [tex]F-22=7.4\times 1.05[/tex]

[tex]F-22=7.4\times 1.05[/tex]

We know that work done is given by

[tex]W=Fs=29.77\times 2=59.54J[/tex]

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