# A piano has been pushed to the top of the ramp at the back of a moving van. The workers think it is safe, but as they walk away, it begins to roll down the ramp. If the back of the truck is 1.0 m above the ground and the ramp is inclined at 20°, how much time do the workers have to get to the piano before it reaches the bottom of the ramp?

This article aims to find the time it takes the workers to reach the piano before it reaches the bottom of the ramp. This article uses the concept of determining the acceleration due to gravity and the length of the ramp. Gravitational acceleration is the acceleration gained by an object due to the force of gravity. Its SI unit is $\dfrac{m}{s ^ { 2 }}$. It has both magnitude and the direction, so it is a vector quantity. Gravitational acceleration is represented by $g$. The standard value of $g$ on the earth’s surface at sea level is $9.8\dfrac {m}{s ^ { 2 }}$.

Step 1

Given values

$h = 1.0 m$

$\theta = 20 ^ { \circ }$

$g = 9.81 \dfrac{ m } { s ^ { 2 } }$

Step 2

When the piano starts moving down the ramp, the gravitational acceleration is:

$a = g \sin \theta$

If we substitute the values into the above equation, we get the desired acceleration value:

$a = ( 9.81 \dfrac {m}{ s ^{2}})( \sin ( 20 ^ { \circ } ))$

$a = ( 9.81 \dfrac{ m }{ s ^ { 2 }} )( 0.34202 )$

$a = 3.35 \dfrac{m}{s ^ { 2 }}$

Length of the ramp is given as:

$\sin \theta = \dfrac {h}{\Delta x}$

$\Delta x = \dfrac{h}{\sin\theta}$

$\Delta x = \dfrac{1.0}{\sin (20^{\circ})}$

$\Delta x = \dfrac{1.0}{0.34202}$

$\Delta x = 2.92m$

So the time for the piano to reach the ground is:

$t = \sqrt {\dfrac{\Delta x}{a}}$

$t = \sqrt {\dfrac{2.92m}{3.35 \dfrac{m}{s^{2}}}}$

$t = 1.32 s$

The time is $1.32s$.

## Numerical Result

The time it takes the workers to reach the piano before it reaches the bottom of the ramp is $1.32 s$.

## Example

The piano was pushed to the top of the ramp in the back of the moving van. The workers think it’s safe, but as they leave, it starts rolling down the ramp. If the back of the truck is $2.0\: m$ above the ground and the ramp is inclined $30^{\circ}$, how much time will the workers take to get to the piano before it reaches the bottom of the ramp?

Solution

Step 1

Given values

$h = 2.0 m$

$\theta = 30^ {\circ}$

$g = 9.81 \dfrac{m}{s^{2}}$

Step 2

When the piano starts moving down the ramp, the gravitational acceleration is:

$a = g \sin \theta$

If we substitute the values into the above equation, we get the desired acceleration value:

$a = (9.81 \dfrac{m}{s^{2}} )(\sin(30^ {\circ}))$

$a = (9.81 \dfrac{m}{s^{2}} )(0.5)$

$a = 19.62 \dfrac{m}{s^{2}}$

Length of the ramp is given as:

$\sin \theta = \dfrac{h}{\Delta x}$

$\Delta x = \dfrac{h}{\sin \theta }$

$\Delta x = \dfrac{2.0}{\sin (30^{\circ})}$

$\Delta x = \dfrac{1.0}{0.5}$

$\Delta x = 4m$

So the time for the piano to reach the ground is:

$t = \sqrt {\dfrac{\Delta x}{a}}$

$t = \sqrt {\dfrac{4m}{19.62 \dfrac{m}{s^{2}}}}$

$t = 0.203 s$

The time is $0.203s$.