# Answer in Physics for Monica #163419

A 250 N force pushes a 25-kg box for 100.0 m. What work is done (by the push only) on the box?

If the box starts from rest and a 50 N frictional force resists the motion, what is the final speed of the box?

If the trup is up a 37 degree incline, what speed will it have after the 100 m?

1) The work is defined as follows:

“W = Fs”

where “F = 250N” is the pushing force, and “s = 100m” is the displacement. Thus:

“W = 250Ncdot 100m = 25times 10^{3}J = 25 space kJ”

2) According to the energy-work theorem, the increase in the kinetic energy of the box is equal to the work of the net force acted on this box:

“K_2-K_1 = F_{net}s”

where “K_1 = 0” is the initial kinetic energy (since the box starts from rest), “F_{net} = 250N-50N = 200N” is the net foce acted on the box (pushing force minus resistance force), and

“K_2 = dfrac{mv^2}{2}”

is the final kinetic energy, where “m = 25kg” is the mass, and “v” is the final speed of the box. Substituting and expressing “v”, obtain:

“dfrac{mv^2}{2}-0 = F_{net}s\nv = sqrt{dfrac{2F_{net}s}{m}}\nv = sqrt{dfrac{2cdot 200cdot 100}{25}} = 40m/s”

3) If the trup is up a “theta =37degree” incline and 100 m counts along it, the projection of the pushing force that actually pushes the box is:

“F = 250cdot cos37degree approx 199.66N”

Then the net force is:

“F_{net} = 199.66N-50N = 149.66N”

and the speed will be:

“v = sqrt{dfrac{2cdot 149.66cdot 100}{25}} approx 34.6m/s”

Answer. 1) 25 kJ, 2) 40 m/s, 3) 34.6 m/s.

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