A cubic block initially at rest slides with a constant acceleration on an inclined plane 10m long in 2s. Calculate the acceleration of the block.
s = 1/2 at^2
so,
1/2 * a * 2^2 = 10
a = 5 m/s^2
assuming, of course, that the block slid the entire length of the plane ...
givens=vi=0m/s,displacement=10m,t=2s>>
>>average acceleration=average velocity/time interval
>>average velocity=diplacement/time interval
>>average velocity=10m/2s=5m/s
>>average acceleration=5/2=2.5m/s^2
i hope this helps :)
Hala - acceleration = total change in velocity/time, not average
a = 10 /2 = 5 m/s^2
what is in your mind I think is total distance = average speed *time
To calculate the acceleration of the block, we need to use the kinematic equation for motion along an inclined plane.
The equation we'll use is:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- s is the distance traveled by the block
- u is the initial velocity (in this case, it is at rest, so u=0)
- t is the time taken (2s)
- a is the acceleration we want to find
From the problem, we know that the distance traveled by the block is 10m and the time taken is 2s.
Plugging in the known values, we have:
\[ 10 = 0 + \frac{1}{2} a (2)^2 \]
Simplifying the equation gives us:
\[ 10 = 2a \]
To isolate the acceleration (a), divide both sides of the equation by 2:
\[ a = \frac{10}{2} \]
Therefore, the acceleration of the block is:
\[ a = 5 \text{ m/s}^2 \]