Answer to Question #168850 in Mechanics | Relativity for April

Question

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.5 m/s the pulling

force is 100 N parallel to the incline, which makes an angle of 20° with the horizontal. The

coefficient of kinetic friction 1s 0.4, and the crate is pulled 5.0 m. (a) How much work is done by the

gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline

system owing to friction. (c) How much work is done by the 100-N force on the crate? (e) What is the

speed of the crate after being pulled 5.0 m

Answer 1

(a) We can find the work done by the gravitational force on the crate from the formula:

$$W_g=mgh.$$

We can find $h$ from the geometry:

$$sin\theta=\dfrac{h}{d},$$ $$h=dsin\theta.$$

Finally, we can calculate the work done by the gravitational force on the crate:

$$W_g=mgdsin\theta=10\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot5.0\ m\cdot sin20^{\circ}=167.6\ J.$$

(b) Let's first find the friction force that acts on the crate:

$$F_{fr}=\mu_kN=\mu_kmgcos\theta,$$ $$F_{fr}=0.4\cdot10\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot cos20^{\circ}=36.83\ N.$$

Then, the increase in internal energy of the crate–incline system due to friction will be:

$$\Delta E_{int}=W_{fr}=F_{fr}d=36.83\ N\cdot5.0\ m=184.15\ J.$$

(c) By the definition of the work done we have:

$$W_{pull}=F_{pull}d=100\ N\cdot5\ m=500\ J.$$

(e) Let's first find the change in the kinetic energy of the crate:

$$\Delta KE=W_{pull}-W_{fr}-W_g,$$ $$\Delta KE=500\ J-184.15\ J-167.6\ J=148.25\ J.$$

Finally, we can find the speed of the crate after being pulled 5.0 m:

$$\Delta KE=\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_f^2,$$ $$v_f=\sqrt{\dfrac{2\Delta KE}{m}+v_i^2},$$ $$v_f=\sqrt{\dfrac{2\cdot148.25\ J}{10\ kg}+(1.5\ \dfrac{m}{s})^2}=5.65\ \dfrac{m}{s}.$$