Question

A 10 kg block slides down a frictionless incline from a height of 5 meters. Assuming the gravitational constant g=9.8 m/s\(^2\), calculate the kinetic energy of the block at the bottom of the incline.

• A. 490 J
• B. 980 J
• C. 4,900 J
• D. 9,800 J

Hint:

Conservation of Mechanical Energy. If only conservative forces (like gravity, ideal springs) do work (i.e., no friction or non-conservative forces), then \(PE_{initial + KE_{initial = PE_{final + KE_{final\). For gravity, \(PE = mgh\).

Answer 1

The Correct Option is A

We can use the principle of conservation of mechanical energy since the incline is frictionless (no energy loss due to friction).
The total mechanical energy (Potential Energy + Kinetic Energy) remains constant.
Let the top of the incline be point 1 and the bottom be point (2) At the top (point 1): - Height \(h_1 = 5\) m.
- Assuming the block starts from rest, initial velocity \(v_1 = 0\), so initial Kinetic Energy \(KE_1 = 0\).
- Potential Energy \(PE_1 = mgh_1\).
At the bottom (point 2): - Height \(h_2 = 0\) (taking the bottom as the reference level).
- Potential Energy \(PE_2 = mgh_2 = 0\).
- Let the final velocity be \(v_2\).
Final Kinetic Energy \(KE_2 = \frac{1}{2}mv_2^2\).
Conservation of energy: \( PE_1 + KE_1 = PE_2 + KE_2 \) $$ mgh_1 + 0 = 0 + KE_2 $$ $$ KE_2 = mgh_1 $$ Substitute the given values: Mass \(m = 10\) kg.
Gravity \(g = 9.
8\) m/s\(^2\).
Height \(h_1 = 5\) m.
$$ KE_2 = (10 \, \text{kg}) \times (9.
8 \, \text{m/s}^2) \times (5 \, \text{m}) $$ $$ KE_2 = 10 \times 49 \, \text{J} = 490 \, \text{J} $$ The kinetic energy of the block at the bottom is 490 J.