A force, varying with distance as shown in the graph above, acts on a 5.5 kg obj

Question

A force, varying  with distance as shown in the graph above, acts on a 5.5 kg object.

A) When is 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 objext from 0 m to 6.0 m. Explain you calculation please!

C) Assuming that there is no friction or air resistance what is the total change i the objects energy? Explain please!

D) What is the velocity of the object when is reaches 6.0 m, assuming the velocity at zero meters is zero? Show work

E) What will be the final velocity is the initial velocity at zero meters is 5 m/s? Show claculation

A force, varying with distance as shown in the graph above, acts on a 5.5 kg object. When is most work done, in the first 4 meters or between 4 and 6 meters? Explain Find the work done by the force in moving the objext from 0 m to 6.0 m. Explain you calculation please! Assuming that there is no friction or air resistance what is the total change i the objects energy? Explain please! What is the velocity of the object when is reaches 6.0 m, assuming the velocity at zero meters is zero? Show work What will be the final velocity is the initial velocity at zero meters is 5 m/s? Show claculation

Explanation / Answer

a). work done in 1st 4m = area under the curve upto 4m = 2*4 = 8 J work done b/w 4 to 6m = (6-4)*3 = 6 J work done in 1st 4 m is more. b). Total work done = 8 + 6 = 14 J c). work done = change in K.E. => change in K.E. = 14 J d). change in K.E. = 14 J => 0.5*5.5*(v^2 - 0) =14 => v = 2.256 m/s e). change in K.E. = 14 J => 0.5*5.5*(v^2 - 5^2) = 14 => v = 5.486 m/s

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