Question

a 6.77 kg block with a specific heat capacity of 1470 j/kg°c slides down a ramp with an angle of inclination of 31.1°. if it starts at the top of the ramp which is 13.3 m long. what is the temperature change of the block if it is traveling at 5.88 m/s at the bottom of the ramp?

(a) 0.340°c
(b) 3.40°c
(c) 0.00340°c
(d) 0.0340°c

Explanation:

Identify the given parameters and physical setup

We are given:

• Mass of the block, \(m = 6.77\text{ kg}\)
• Specific heat capacity, \(c = 1470\text{ J/(kg}\cdot^\circ\text{C)}\)
• Angle of inclination, \(\theta = 31.1^\circ\)
• Length of the ramp, \(L = 13.3\text{ m}\)
• Final velocity at the bottom, \(v = 5.88\text{ m/s}\)
• Acceleration due to gravity, \(g \approx 9.8\text{ m/s}^2\) (or \(9.81\text{ m/s}^2\))

We need to find the temperature change \(\Delta T\) of the block, assuming all the mechanical energy lost to friction is converted into thermal energy absorbed entirely by the block.

Calculate the initial mechanical energy

Using the Conservation of Energy knowledge point

$$
LATEXBLOCK0
$$

Calculate the final kinetic energy

Using the Conservation of Energy knowledge point

$$ E_f = \frac{1}{2}mv^2 = 0.5 \cdot 6.77 \cdot (5.88)^2 \approx 0.5 \cdot 6.77 \cdot 34.5744 \approx 117.04\text{ J} $$

Determine the thermal energy generated

Using the Conservation of Energy knowledge point

$$ Q = E_i - E_f = 455.79 - 117.04 = 338.75\text{ J} $$

Calculate the temperature change of the block

Using the Specific Heat Capacity and Heat Transfer Equation knowledge points

$$
LATEXBLOCK1
$$

Answer:

• 0.340°C
• 3.40°C
• 0.00340°C
• 0.0340°C (Correct answer)