A block is released from rest on an inclined plane and moves 3.2 m during the next 3.9 s. The acceleration of gravity is 9.8 m/s2 .
What is the magnitude of the acceleration of the block? Answer in units of m/s2.
The block is 12kg and the incline is 36 degrees.
find the acceleration down the block:
vf^2=1/2 acceleration*distance
vf=2*avgvelociyt=2*3.2/3.9
putting this together, solve for acceleaetion
a= 2*(2*3.2/3.9)^2 /3.9
I tried this problem with a=2(l)/t^2, a=2(3.2)/3.95^2 = 0.41019 and the answer is not correct and I am not sure what is wrong with it or how to do it.
Can someone please walk me through what equation I need to use to do this?
post it.
Well, well, well! It seems like this block is having quite the adventure on this inclined plane. Let's see if we can help it out.
To find the magnitude of the acceleration, we can use the equation of motion for objects moving on inclined planes. The equation is:
a = g * sin(θ)
where a is the acceleration, g is the acceleration due to gravity, and θ is the angle of the incline.
In this case, g is 9.8 m/s^2 and θ is 36 degrees. So, let's plug in those values and calculate the magnitude of the acceleration:
a = 9.8 m/s^2 * sin(36 degrees)
Using my handy-dandy calculator, the magnitude of the acceleration is approximately 5.94 m/s^2.
So, there you have it! The magnitude of the acceleration of the block is 5.94 m/s^2. Now, I hope this block enjoyed its little journey down the incline. Maybe it can put on a show for us next time!
To find the magnitude of the acceleration of the block, we can use the kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
where:
- s is the displacement (3.2 m)
- u is the initial velocity of the block (0 m/s as it is released from rest)
- a is the acceleration of the block (to be determined)
- t is the time (3.9 s)
We can rearrange the equation to solve for a:
\[ a = \frac{2(s - ut)}{t^2} \]
Substituting the given values:
\[ a = \frac{2(3.2 - 0 \cdot 3.9)}{3.9^2} \]
Now, let's calculate the value:
\[ a = \frac{6.4}{15.21} \approx 0.421 \, \text{m/s}^2 \]
Therefore, the magnitude of the acceleration of the block is approximately 0.421 m/s².
h = 3.2*sin36 = 1.88 m. Ht. of incline.
V^2 = Vo^2 + 2g*h = 0 + 19.6*1.88 = 36.9
V = 6.1 m/s = Final velocity.
V^2 = Vo^2 + 2a*d.
6.1^2 = 0 + 2*a*3.2,
a = 5.8 m/s^2.
Another Method:
M*g = 12*9.8 = 117.6 N. = Wt. of block.
Fp = 117.6*sin36 = 69.1 N. = Force parallel to the incline.
Fp = M*a.
69.1 = 12*a,
a = 5.8 m/s^2.