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a 9 kilogram box slides down a frictionless inclined plane a distance o…

Question

a 9 kilogram box slides down a frictionless inclined plane a distance of 10 meters. if the object starts from rest, what is the kinetic energy of the box at the bottom of the incline? (g = 10 m/s²)

Explanation:

Step1: Identify Energy Conservation

Since the plane is frictionless, mechanical energy is conserved. Initial kinetic energy (\(KE_i\)) is 0 (starts from rest), so final kinetic energy (\(KE_f\)) equals initial potential energy (\(PE_i\)).

Step2: Calculate Potential Energy

Potential energy formula: \(PE = mgh\), where \(m = 9\,\text{kg}\), \(g = 10\,\text{m/s}^2\), \(h = 7\,\text{m}\) (height of incline).
Substitute values: \(PE = 9 \times 10 \times 7\).

Step3: Compute the Value

\(9 \times 10 \times 7 = 630\,\text{J}\). Thus, \(KE_f = PE_i = 630\,\text{J}\).

Answer:

630 J (corresponding to the option "630 J")