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 is represented by the formula:

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

where:

- \( F \) is the force applied,
- \( m \) is the mass of the object,
- \( a \) is the acceleration of the object.

Denise has provided the following values:
- Mass (\( m \)) of the wooden block is \( 0.5 \) kilograms.
- Force (\( F \)) measured by the sensor is \( 4.9 \) newtons.

We need to solve for the acceleration \( a \). To isolate \( a \) in the formula, we rearrange it as follows:

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

Substitute the given values into the equation:

[tex]\[ a = \frac{4.9 \, \text{N}}{0.5 \, \text{kg}} \][/tex]

Now we can calculate the acceleration:

[tex]\[ a = 9.8 \, \text{m/s}^2 \][/tex]

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

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