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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]


Sagot :

To find the acceleration of the wooden block when it hits the sensor, we can use Newton's second law of motion, which states that force ([tex]\( F \)[/tex]) equals mass ([tex]\( m \)[/tex]) multiplied by acceleration ([tex]\( a \)[/tex]). This can be written as:

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

Given the values:
- The mass ([tex]\( m \)[/tex]) of the wooden block is 0.5 kilograms.
- The force ([tex]\( F \)[/tex]) measured by the sensor is 4.9 newtons.

We need to solve for the acceleration ([tex]\( a \)[/tex]). We can rearrange the equation to solve for [tex]\( a \)[/tex]:

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

Substituting in the known values:

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

When we perform the division:

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

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

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