A force, varying with distance, acts on a 5.5 kg object.In the first 4 meters, i

Question

A force, varying with distance, acts on a 5.5 kg object.In the first 4 meters, it has a constant force of 2.0 N. Between 4.0 and 6.0 m, there is a constant force of 3.0 N.

a) When is the most work done, in the first 4 meters or between 4 and 6 meters? Explain

b) Find the work done by the force in moving the object from 0 m to6.0 m. Explain calculation.

c) Assuming that there is no friction or air resistance, what is the total cange in the object's energy? Explain.

d) What is the velocity of the object when it reaches 6.0 m, assuming the velocity at 0 m = 0 m/s? Show calculation,

e) What will be the final velocity if the initial velocity at 0 m is 5 m/s? Don't jump to a conclusion! (Not really sure what that is about). Show calculation.

Thanks to anyone who answers. I attempted to upload the graph, but for some reason it is not working.

Explanation / Answer

(a) work done in 4m = 2 x 4 =8 J

work done in 4mto 6m = 3 x 2 = 6 J

work done in first 4 m will be more

(b) total work done = 8 + 6 = 14 j

(c)total change in obect energy = work done on object = 14 J

(d) 0.5mv^2 = 14

v = 2.25 m/s

(e) initial energy of object + work done = final energy

0.5 x 5.5 x 5^2 + 14 = 0.5mv^2

v = 5.48 m/s

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