HELP!!
Two blocks of mass m1 = 320 g and m2 = 725 g are pushed by a force. The coefficient of kinetic friction between each block and the ground is 0.4. The blocks move together (with one pushing the other) with an acceleration of 225 cm/s2. What is the value of the applied force in Newtons? Hint: Make sure to include the frictional force for each block.
320+725 = 1045 g = 1.045 kg
friction force = 1.045*9.81 * .4
F - friction force = 1.045 * 2.25 m/s^2
To find the value of the applied force, we need to consider the forces acting on both blocks individually. Let's break it down step by step:
Step 1: Calculate the frictional force on each block.
The frictional force can be calculated using the formula: Frictional force = coefficient of friction * Normal force.
Since both blocks move together, the normal force acting on each block is equal to its weight (mg).
For the first block (m1 = 320 g = 0.32 kg):
Frictional force1 = coefficient of friction * Normal force1
= 0.4 * (0.32 kg * 9.8 m/s^2) [converting to SI units]
= 1.23 N
For the second block (m2 = 725 g = 0.725 kg):
Frictional force2 = coefficient of friction * Normal force2
= 0.4 * (0.725 kg * 9.8 m/s^2) [converting to SI units]
= 2.83 N
Step 2: Calculate the net force acting on both blocks.
The net force can be obtained using Newton's second law of motion: Net force = mass * acceleration.
For both blocks combined, since they move together with the same acceleration:
Net force = (m1 + m2) * acceleration
= (0.32 kg + 0.725 kg) * 225 cm/s^2 [converting acceleration to m/s^2]
= 196.45 N
Step 3: Calculate the applied force.
The applied force should overcome the sum of the frictional forces and provide the net force:
Applied force = Net force + Frictional force1 + Frictional force2
= 196.45 N + 1.23 N + 2.83 N
= 200.51 N
Therefore, the value of the applied force is 200.51 Newtons.