A,horizontial,force,of,40N,actingon,a,block,of,frictionless,level,surface,produces,an,acceleration,of,2.5m/s^2.A,second,block,with,a,mass,of,4.0kg,is,dropped,onto,the,first,.What,is,the,magnitude,of,acceleration,of,the,combination,of,blocks,if,the,same,force,continues,to,act?,(assume,that,the,second,block,does,not,slide,on,the,first,block.)
Would it be asking to much for you to use spaces instead of commas between words?
Get the mass M1 of the first block from
F = 40 N = M1* 2.5 m/s^2
M1 = 16 kg
Dropping the second block on top increases total mass to 20 kg.
With the same force acting, acceleration is reduced by a factor 16/20 = 0.8
To find the magnitude of acceleration of the combination of blocks, we need to consider the net force acting on the system.
First, let's calculate the mass of the first block. Since we are given the net force and the acceleration of the first block, we can use Newton's second law:
Force = mass * acceleration
40N = mass * 2.5m/s^2
Rearranging the equation to solve for mass, we have:
mass = 40N / 2.5m/s^2 = 16 kg
Since the net force acts on both blocks together, we can assume that the force is also acting on the second block.
Next, we calculate the net force on the system. Since the second block is dropped onto the first block, there is an increase in mass but no additional force is applied. Therefore, the net force acting on the system remains 40N.
To find the total mass of the combined system, we add the mass of both blocks:
Total mass = mass of the first block + mass of the second block
Total mass = 16kg + 4.0kg = 20kg
Now, we can use Newton's second law again to find the acceleration of the combination of blocks:
Force = mass * acceleration
40N = 20kg * acceleration
Rearranging the equation to solve for acceleration, we have:
acceleration = 40N / 20kg = 2.0 m/s^2
Therefore, the magnitude of acceleration of the combination of blocks when the same force continues to act is 2.0 m/s^2.