Question

A 120 kg box slides from rest down to the bottom of a 30º incline with a 100 m long slope. The coefficient of kinetic friction between the box and the inclined surface is 0.25.

a) How much energy is transferred to thermal energy by friction during the slide?

b) Find the box's speed as it reaches the bottom of the incline.

Answer 1

a) The energy transferred to thermal energy by friction during the slide is 25,487.13 J.

b) The box's speed as it reaches the bottom of the incline is 23.58 m/s.

Step-by-step explanation:

Initial Potential Energy:

Mass of the box (m) = 120 kg

Angle of the incline (θ) = 30°

Length of the incline (l) = 100 m

Acceleration due to gravity (g) = 9.81 m/s²

Potential energy (PE) is calculated as:

PE = mghsinθ

= 120 kg * 9.81 m/s² * 100 m * sin(30°)

= 58,699.82 J

Work Done by Friction:

Coefficient of kinetic friction (μ) = 0.25

Work done by friction (W_f) is calculated as:

W_f = μmgcosθl

= 0.25 * 120 kg * 9.81 m/s² * cos(30°) * 100 m

= 25,487.13 J

Final Kinetic Energy:

Total energy is conserved (PEi = KEf + W_f)

Final kinetic energy (KEf) is calculated as:

KEf = PEi - W_f

= 58,699.82 J - 25,487.13 J

= 33,212.69 J

Box's Speed:

Speed (v) is calculated using the formula for kinetic energy:

KEf = 1/2 mv²

v = √(2KEf / m)

= √(2 * 33,212.69 J / 120 kg)

= 23.58 m/s