Aperson 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?

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

Balance force is given by

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

So

We know that work done is given by

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

compass can be used day and night, cloudy or sunny