Question

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?

Answer 1

1) The work is defined as follows:

$$W = Fs$$

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

$$W = 250N\cdot 100m = 25\times 10^{3}J = 25 \space kJ$$

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:

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

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

$$F = 250\cdot \cos37\degree \approx 199.66N$$

Then the net force is:

$$F_{net} = 199.66N-50N = 149.66N$$

and the speed will be:

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

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