Question

A 1200 crate slides 11 down a ramp that makes an angle of 35 with the horizontal.

If the crate slides at a constant speed, how much thermal energy is created?

Answer 1

Concepts and reason

The concepts required solve this problem are law of conservation of energy and resolution of a force.

Initially, draw a useful diagram to find the height of the ramp and the component of the weight of the crate along the inclined plane of the ramp. Later, write the expression for the law of conservation of energy. After that, write the expression for kinetic energy and discuss how the kinetic energy is constant if the speed of the crate is constant. Further, write the expression for the potential energy and used this in expression for law of conservation of energy and solve for work done by the non-conservative forces.

Finally, discuss how the thermal energy relates with the work done by the non-conservative forces and find the thermal energy created by the crate.

Fundamentals

From the law of conservation of energy, the energy neither be created nor be destroyed. The energy can be transferred from one form of energy into other form of energy. In other words, the mechanical energy of an isolated system is constant if there are any non-conservative forces are involved. The mechanical energy of a system is equal to the sum of kinetic energy and potential energy.

The expression for the law of conservation of energy if non-conservative forces are involved in the system is,

${U_i} + {K_i} + {W_{{\rm{nc}}}} = {U_f} + {K_f}$

Here, ${U_i}$ is the initial potential energy, ${K_i}$ is the initial kinetic energy, ${W_{{\rm{nc}}}}$ is the work done by the non-conservative forces, ${U_f}$ is the final potential energy, and ${K_f}$ is the final kinetic energy.

The gravitational potential energy of an object which is at a height h above a reference level can be calculated using the following formula:

$U = mgh$

Here, m is the mass of the object and g is the acceleration due to gravity.

The expression for the kinetic energy of an object is as follows:

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

Here, m is the mass of the object and v is the speed of the object.

In the question, there are no units are given for the parameters. I have considered the units of all the given parameters in SI system of units and the result is also provided in SI system of unit.

The following figure represents the useful diagram to find the height of the ramp and the component of the weight of the crate along the inclined plane of the ramp:

In the figure, L represents the length of the ramp and $\theta$ is the angle of inclination of the ramp.

The kinetic energy of the crate changes if its speed is changed. The initial and final kinetic energy of the crate will be same as it is sliding with constant speed.

${K_i} = {K_f}$

From the figure, $\sin \theta$ can be written as follows:

$\sin \theta = \frac{h}{L}$

Rearrange the equation for h.

$h = L\sin \theta$

The expression for the gravitational potential energy is,

$U = mgh$

Substitute $L\sin \theta$ for h, and derive the expression for the potential energy of the crate at top of the ramp (with respect to the reference level at bottom of the ramp) that is initial potential energy of the crate.

${U_i} = mg\left( {L\sin \theta } \right)$

Substitute 0 for h, and derive the expression for the potential energy of the crate at bottom of the ramp (with respect to the reference level at bottom of the ramp) that is final potential energy of the crate.

$\begin{array}{c}\\{U_f} = mg\left( 0 \right)\\\\ = 0\\\end{array}$

The expression for the law of conservation of energy if non-conservative forces are involved in the system is,

${U_i} + {K_i} + {W_{{\rm{nc}}}} = {U_f} + {K_f}$

Substitute 0 for ${U_f}$ , $mg\left( {L\sin \theta } \right)$ for ${U_i}$ , ${K_i}$ for ${K_f}$ , and solve for ${W_{{\rm{nc}}}}$ .

$\begin{array}{c}\\mg\left( {L\sin \theta } \right) + {K_i} + {W_{{\rm{nc}}}} = 0 + {K_i}\\\\{W_{{\rm{nc}}}} = - mgL\sin \theta \\\end{array}$

The expression for the work done by the non-conservative forces is,

${W_{{\rm{nc}}}} = - mgL\sin \theta$

Substitute 1200 N for mg, 11 m for L, and $35^\circ$ for $\theta$ .

$\begin{array}{c}\\{W_{{\rm{nc}}}} = - \left( {1200{\rm{ N}}} \right)\left( {11{\rm{ m}}} \right)\sin 35^\circ \\\\ = - {\rm{7}}{\rm{.57}} \times {10^3}{\rm{ J}}\\\end{array}$

Therefore, the amount of work done by the non-conservative forces acting on the crate is ${\rm{7}}{\rm{.57}} \times {10^3}{\rm{ J}}$ .

The work done by the non-conservative forces acting on the crate converts in the thermal energy. Thus, the thermal energy created by the crate during its slide on the ramp is ${\rm{7}}{\rm{.57}} \times {10^3}{\rm{ J}}$ .

Ans:

The thermal energy created by the crate during its slide on the ramp is ${\rm{7}}{\rm{.57}} \times {10^3}{\rm{ J}}$ .