Question

A small child with a mass of 20.0 kg slides down a 7.0 m frictionless slide that has a vertical height of 4.5 m. What is her speed at the bottom of the slide? 90.0 m/s9.4 m/s4.4 m/s16.0 m/s

Answer 1

According to problem

mass(m)= 20.0kg;

Displacement= s

Here

$\sin\theta=\frac{4.5}{7}$

Now component of mg parallel to inclined plane = mg sinθ is responsible for acceleration.

Now we can write

$\begin{gathered} ma=\text{ mg}\sin\theta; \\ 20a=\text{ 20}\times9.8\times\frac{4.5}{7}\text{ :} \\ a=\text{ 6.3 m/s}^2 \end{gathered}$

Using equation of motion

$\begin{gathered} v^2=u^2+\text{ 2as;} \\ v^2=\text{ o+ 2}\times6.3\times7.0=\text{ 88.2;} \\ v=\sqrt{88.2}\text{ = 9.4 m/s \lparen approx\rparen} \end{gathered}$

Final answer is 9.4m/s