Q33E - University Physics with Modern Physics

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 travelled 6.80 m to the bottom of the incline is 3.80 m/s. What is the speed of the block when it is 3.40 m from the top of the incline?

Step-by-Step Solution

Answer

Speed when travelling is top of the incline= 2.687 m/s.

Step 1: Identification of the given data

• Speed when travelling is bottom of the incline=3.80 m/s.
• Distance travelled = 6.80 m.

Step 2: calculation for the speed

Total speed is calculated by using ,

${\mathrm{v}}^2={\mathrm{u}}^2+(2\times \mathrm{a}\times \mathrm{s})$ 

Where, v= final velocity, u=initial velocity, a=acceleration due to gravity, and s=distance travelled

Substituting the values ,

$$\begin{gathered} 3.8^2=0+(2\times \mathrm{a}\times 6.80) \\ \mathrm{a}=0.941 \end{gathered}$$ 

Now using the value of a for calculation of speed when distance travelled is 3.40 m,

Substituting the given values,

$$\begin{gathered} {\mathrm{v}}^2=0+(2\times 0.941\times 3.40) \\ {\mathrm{v}}^2=6.3988 \\ \mathrm{v}=\sqrt{6.3988} \\ \mathrm{v}=2.687 \mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the final speed of the block is $\mathrm{v}=2.687 \mathrm{m}/\mathrm{s}$