Select the correct answer.

Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop. She drops a wooden block with a mass of 0.5 kilograms on a sensor on the floor. The sensor measures the force of the impact as 4.9 newtons. What is the acceleration of the wooden block when it hits the sensor? Use [tex]$F = ma$[/tex].

A. [tex]$2.45 \, m/s^2$[/tex]
B. [tex]$4.4 \, m/s^2$[/tex]
C. [tex]$5.4 \, m/s^2$[/tex]
D. [tex]$9.8 \, m/s^2$[/tex]



Answer :

To determine the acceleration of the wooden block when it hits the sensor, we can use Newton's second law of motion, which states that the force \( F \) acting on an object is equal to the mass \( m \) of the object times its acceleration \( a \). This relationship is expressed by the formula:

[tex]\[ F = m \times a \][/tex]

Given the values:
- The mass of the wooden block, \( m = 0.5 \) kilograms
- The force measured by the sensor, \( F = 4.9 \) newtons

We need to find the acceleration \( a \). Rearrange the formula to solve for \( a \):

[tex]\[ a = \frac{F}{m} \][/tex]

Substitute the known values into this formula:

[tex]\[ a = \frac{4.9 \text{ newtons}}{0.5 \text{ kilograms}} \][/tex]

Calculate the result:

[tex]\[ a = 9.8 \, \frac{\text{meters}}{\text{second}^2} \][/tex]

Therefore, the acceleration of the wooden block when it hits the sensor is \( 9.8 \, \frac{\text{m}}{\text{s}^2} \).

So, the correct answer is:
D. [tex]\( 9.8 \, \frac{\text{m}}{\text{s}^2} \)[/tex]