Answer to Question #168173 in Physics for van

Question

A huge box having a mass of 6.80 kg is pulled down along a frictionless surface inclined at 42.0° below the horizontal by a constant force of 7.21 N that is parallel to the incline. The baggage was initially at rest at the top of the ramp and is moving at 5.52 m/s when it reaches the bottom of the ramp. The coefficient of kinetic friction between the box and the inclined surface is 0.39. (a) How much work is done on the box by all forces acting on it? (b) How long is the inclined surface?

Answer 1

According to Newton's second law:

$$Ox: mg\text{ sin}\theta +F-\mu N=ma,\\ Oy: -mg\text{ cos}\theta+N=0.$$

Find the acceleration:

$$a= g\text{ sin}\theta +\frac Fm-\mu g\text{ cos}\theta=4.78\text{ m/s}^2.$$

Find the length of the incline:

$$L=\frac{v_f^2-v_i^2}{2a}=\frac{5.52^2-0}{2·4.78}=3.19\text{ m}.$$

Energy of the box on top of the incline:

$$E_1=mgh=6.8·9.8·3.19\text{ sin}42º=142\text{ J}.$$

Energy on the bottom consists of the kinetic energy:

$$E_2=\frac12 mv^2$$

Work is force times distance. For the force F:

$$W_F=FL=7.21·3.19=23.0\text{ J}.$$

For the force of friction (it works in opposite direction compared to F):

$$W_f=-fL=-\mu N L=-61.6\text{ J}.$$

Work done by the gravity (it works in the same direction as F):

$$W_g=Lmg\text{ sin}\theta =142\text{ J}.$$

Work done by all forces:

$$W=W_F+W_f+W_g=103\text{ J}.$$