Question
How to find the magnitude of acceleration of a 2.00 kg block sliding down a 25.0 degree incline, moving 1.80 m in 1.20 s?
Answers
GPT-4o mini
To find the magnitude of acceleration of a block sliding down an incline, we can use the equation of motion that relates the distance traveled, the time taken, initial velocity, and acceleration. The basic equation of motion is:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Where:
- \(d\) = distance traveled (1.80 m)
- \(v_0\) = initial velocity (we'll assume it's 0 if the block starts from rest)
- \(t\) = time (1.20 s)
- \(a\) = acceleration (what we want to find)
Assuming the block starts from rest, \(v_0 = 0\). Thus, the equation simplifies to:
\[
d = \frac{1}{2} a t^2
\]
Substituting the values we have:
\[
1.80 = \frac{1}{2} a (1.20)^2
\]
Calculating \( (1.20)^2 = 1.44 \):
\[
1.80 = \frac{1}{2} a \cdot 1.44
\]
Multiplying both sides by 2:
\[
3.60 = a \cdot 1.44
\]
Now, solve for acceleration \(a\):
\[
a = \frac{3.60}{1.44}
\]
Calculating this will give:
\[
a \approx 2.50 \text{ m/s}^2
\]
Thus, the magnitude of the acceleration of the block sliding down the incline is approximately **2.50 m/s²**.
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Where:
- \(d\) = distance traveled (1.80 m)
- \(v_0\) = initial velocity (we'll assume it's 0 if the block starts from rest)
- \(t\) = time (1.20 s)
- \(a\) = acceleration (what we want to find)
Assuming the block starts from rest, \(v_0 = 0\). Thus, the equation simplifies to:
\[
d = \frac{1}{2} a t^2
\]
Substituting the values we have:
\[
1.80 = \frac{1}{2} a (1.20)^2
\]
Calculating \( (1.20)^2 = 1.44 \):
\[
1.80 = \frac{1}{2} a \cdot 1.44
\]
Multiplying both sides by 2:
\[
3.60 = a \cdot 1.44
\]
Now, solve for acceleration \(a\):
\[
a = \frac{3.60}{1.44}
\]
Calculating this will give:
\[
a \approx 2.50 \text{ m/s}^2
\]
Thus, the magnitude of the acceleration of the block sliding down the incline is approximately **2.50 m/s²**.