Question

a small block has constant acceleration as it slides down a frictionless incline. the block is released from rest at the top of the incline, and its speed after it has traveled 7.80 m to the bottom of the incline is 3.80 m/s. what is the speed of the block when it is 3.80 m from the top of the incline? express your answer with the appropriate units.

Explanation:

Step1: Find acceleration using \( v^2 = u^2 + 2as \)

Given \( u = 0 \, \text{m/s} \), \( s = 7.80 \, \text{m} \), \( v = 3.80 \, \text{m/s} \).
\( a=\frac{v^2 - u^2}{2s}=\frac{(3.80)^2 - 0}{2\times7.80}=\frac{14.44}{15.6}\approx0.9257 \, \text{m/s}^2 \).

Step2: Find speed at \( s' = 3.80 \, \text{m} \)

Use \( v'^2 = u^2 + 2as' \), \( u = 0 \), \( a\approx0.9257 \, \text{m/s}^2 \), \( s' = 3.80 \, \text{m} \).
\( v'=\sqrt{0 + 2\times0.9257\times3.80}=\sqrt{7.0254}\approx2.65 \, \text{m/s} \).

Answer:

\( 2.65 \, \text{m/s} \)