A time varying force F(t) changes with t as shown in (a). The force pushes a 6.0

Question

A time varying force F(t) changes with t as shown in (a). The force pushes a 6.00kg package on a horizontal and frictionless track. Determine the final velocity Vf of the package if its initial is Vi = 4.00m/s. Also find the final KEf of th package at the end of the 12 seconds of pushing it by F(t). At the end of the track a spring with the spring constant of k=4200N/m stops the package. Obtain the compression x of the spring. (Obtain area under the curve c as the impulse ChangeP on the package)

F(N)

16           *(4,16)           *(10,16)

12                                                            ( Actual graph has lines connecting each point)

8

4   *(0,0)                               * (12,0)

0     2     4    6     8     10     12 (T)(s)

F--->[Box]---->Vi         [Box]---->Vf        [spring----------]

                                   [box]---->Vf     [Box][spring--------]

(This is the best I could do with the graph and spring diagram please help m solve this for 5 stars)

Vf=?

KEf=?

X=?

Explanation / Answer

the work done on the package to move it a distance S is

W = F(t) x S

the work done is equal to the change in kinetic energy of the package therefore

W = (1/2)m x (vf^2 - vi^2)

or vf^2 - vi^2 = (2W/m)

or vf = ((2W/m) + vi^2)^1/2

the final kinetic energy is

Kf = (1/2)m x vf^2

the final kinetic energy of the package is equal to the kinetic energy of the spring therefore

(1/2)m x vf^2 = (1/2)k x d^2

or d = (m/k)^1/2 x vf

where d is compression distance

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