Question

At a certain factory, 300 kgcrates are dropped vertically from a packing machine onto a conveyor belt moving at 1.20 m/s(Fig. 8-64). (A motor maintains the belt’s constant speed.) The coefficient of kinetic friction between the belt and each crate is 0.400. After a short time, slipping between the belt and the crate ceases, and the crate then moves along with the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (a) the kinetic energy supplied to the crate, (b) the magnitude of the kinetic frictional force acting on the crate, and (c) the energy supplied by the motor. (d) Explain why answers (a) and (c) differ.

Short answer

1. The value of the kinetic energy K = 216 J
2. The magnitude of the kinetic frictional force$\left|f_k\right|=1.18x{10}^{3}N$
3. The energy supplied by the motor is E = 432 J

d. The difference is because of the thermal energy.

Step-by-step solution

Given

Mass of the crate m = 300 kg

Velocity of the crate v = 1.20 m/s

Coefficient of friction$\mu =0.400$

Determine the concept and the formulas:

The problem is based on the principle of conservation of mechanical energy. According to this principle the energy can neither be created nor be destroyed; it can only be internally converted from one form to another if the forces doing work on the system are conservative in nature. By using the concept of frictional force and mechanical energy of non - conservative systems we can find the required values.

Formula:

$K=\frac{1}{2}\times m{v}^2$

role="math" localid="1661403272907" $$\begin{gathered} f_k={\mu }_{k}N \\ {E}_{mec}=K+W_{friction} \\ {W}_{friction}=f_{k}d \end{gathered}$$

(a) Calculate the kinetic energy supplied to the crate

Kinetic Energy supplied to the crate

$K=\frac{1}{2}\times m{v}^2$

$\Rightarrow K=\frac{1}{2}\times 300\times {\left(1.20\right)}^2$

$\Rightarrow K=\frac{432}{2}$

$\Rightarrow K=216J$

(b) Calculate magnitude of kinetic frictional force

Determine the frictional force experienced on the particle as:

$$\begin{gathered} f_k={\mu }_k\mathrm{N} \\ \Rightarrow {\mathrm{f}}_{\mathrm{k}}=0.4\times \mathrm{m}\times \mathrm{g} \\ \Rightarrow {\mathrm{f}}_{\mathrm{k}}=0.4\times 300\times 9.8 \\ {\mathrm{f}}_{\mathrm{k}}=1176\mathrm{N} \\ \approx 1.18\times {10}^3\mathrm{N} \end{gathered}$$

(c) Energy supplied by the motor

$E=KineticEnegry+workdoneagainstthefriction$

$WorkDoneagainstthefriction=W_f=f_{k}d$

To find value of d,

Acceleration of the crate can be calculated as,

$$\begin{gathered} f_k=maa \\ =\frac{f_k}{m}a \\ =\frac{1176}{300}a \\ =3.92\frac{m}{s^2} \end{gathered}$$

Time taken by the crate to achieve belts velocity is:

$\begin{array}{l}t=\frac{v}{a}t\\ =\frac{1.20}{3.92}t\\ =0.3061sec\end{array}$

Distance covered by the crate during this time,

$\begin{array}{l}d=v\times t\\ =1.20\times 0.3061\\ =0.3673m\end{array}$

Determine the work done by the friction as:

$\Rightarrow W_{friction}=f_{k}d$
$$\begin{gathered} W_{friction}=1176\times 0.3673W_{friction} \\ =432J \end{gathered}$$

(d) Explain why answers (a) and (c) differ

$E_{motor}\ne K$

Because some of the energy supplied by the motor is converted into heat energy due to work done against the friction.