Two blocks of 10kg and 4 kg ,respectively, are in contact with each other on frictionless,horizontal plane.They are moving at 20 metre per second.A 28N force is now applied horizontally to the opposite direction.Draw separate force diagrams for the two blocks, showing all the horizontal forces acting on a blocks,
calculate the acceleration of the blocks(-2metre per seconds squad,calculate the magnitude of the resultant horizontal force acting on the 10kg block (20N),calculate the force that the 4kg block exerts on the 10kg block and give an axplanation for the answer,for how mamy second must 28N retarding force be applied to achieve a 10metre per seconds speed in opposite direction(15seconds )
acceleration=force/mass=28/(10+4)=2m/s^2
force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block
vf=vi+at
-10=20*28/14 * t
t=30/2=15sec
Thanks, please show me all answers
Separate force diagrams for the two blocks, showing all the horizontal forces acting on each block:
10kg block:
|-------->| (20N) |-------->| (28N) |
mg FN1 FN2
4kg block:
|-------->| (28N) |-------->| (20N) |
mg FN1 FN2
Calculating the acceleration of the blocks (-2m/s^2) using Newton's second law:
For the 10kg block:
Sum of the horizontal forces = mass * acceleration
(20N - 28N) = 10kg * acceleration
-8N = 10kg * acceleration
Acceleration = -8N / 10kg
Acceleration = -0.8 m/s^2
For the 4kg block:
Sum of the horizontal forces = mass * acceleration
(28N - 20N) = 4kg * acceleration
8N = 4kg * acceleration
Acceleration = 8N / 4kg
Acceleration = 2 m/s^2
Calculating the magnitude of the resultant horizontal force acting on the 10kg block (20N):
Magnitude of resultant force = mass * acceleration
Magnitude of resultant force = 10kg * (-0.8m/s^2)
Magnitude of resultant force = -8N
Calculating the force that the 4kg block exerts on the 10kg block:
Using Newton's third law: the force exerted by the 4kg block on the 10kg block is equal in magnitude and opposite in direction to the force exerted by the 10kg block on the 4kg block. Therefore, it is 20N.
Calculating for how many seconds the 28N retarding force must be applied to achieve a 10m/s speed in the opposite direction (15 seconds):
Using the equation of motion:
Final velocity = Initial velocity + (acceleration * time)
0m/s = 10m/s + (-2m/s^2 * time)
-10m/s = -2m/s^2 * time
Time = -10m/s / -2m/s^2
Time = 5 seconds
Therefore, the 28N retarding force must be applied for 15 seconds to achieve a 10m/s speed in the opposite direction.
To draw separate force diagrams for the two blocks and calculate the acceleration, we need to consider the forces acting on each block individually.
1. For the 10kg block:
- The gravitational force (weight) acting vertically downward can be represented by a vector pointing downwards with a magnitude of 10kg * 9.8 m/s^2 (acceleration due to gravity).
- The normal force acting perpendicular to the surface can be represented by a vector pointing upwards with the same magnitude as the weight.
- Since the surface is frictionless, there is no horizontal force acting on the block.
2. For the 4kg block:
- The gravitational force (weight) acting vertically downward can be represented by a vector pointing downwards with a magnitude of 4kg * 9.8 m/s^2.
- The normal force acting perpendicular to the surface can be represented by a vector pointing upwards with the same magnitude as the weight.
- Since the surface is frictionless, there is no horizontal force acting on the block.
The fact that the blocks are in contact with each other means that the only horizontal force acting on both blocks is the force applied. Therefore, the force diagram for both blocks would show a force vector of 28N pointing in the opposite direction to their initial motion.
To calculate the acceleration of the blocks, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). Since the force applied is the only horizontal force acting on the blocks, we can equate it to the sum of the masses multiplied by the acceleration (28N = (10kg + 4kg) * a). Solving for acceleration (a), we find a = -2 m/s^2, which is negative because the acceleration is in the opposite direction to the initial motion.
To calculate the magnitude of the resultant horizontal force acting on the 10kg block, we can use the equation F = ma, where F is the net force and a is the acceleration. Since we know the mass of the block (10kg) and the acceleration (-2 m/s^2), we can substitute the values into the equation to find that the magnitude of the resultant force is 20N.
The force that the 4kg block exerts on the 10kg block can be calculated using Newton's third law, which states that for every action, there is an equal and opposite reaction. Since the blocks are in contact and there is no friction, the force exerted by the 4kg block on the 10kg block is equal in magnitude but in the opposite direction to the force exerted by the 10kg block on the 4kg block, which is 20N.
To calculate how many seconds the 28N retarding force must be applied to achieve a 10 m/s speed in the opposite direction, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time. Rearranging the equation to solve for time, we have t = (v - u) / a. Since the initial velocity is 20 m/s (opposite direction to the desired final velocity), the final velocity is 10 m/s, and the acceleration is -2 m/s^2 (negative because it is in the opposite direction), we substitute the values into the equation to find t = (10 m/s - (-20 m/s)) / (-2 m/s^2) = 15 seconds.