A 60 kg block slides along the top of a 100 kg block. The lighter block has an acceleration of 3.7 m/s2 when a horizontal force F= 350 N is applied. Assuming there is no friction between the bottom 100 kg block and the horizontal frictionless surface but there is friction between the blocks. Find the acceleration of the 100 kg block during the time the 60 kg block remains in contact.
friction between the two blocks:
Ff=60*3.7
well, this Ff drives the lower block.
Ff=ma=100*a or
a=60*3.7/100=2.22 acceleaetion of lower block.
so the relative acceleration on the top block=3.7-2.22=1.5m/s^2
to time the blocks are in contact
d=1/2 a t^2 or
t= sqrt(2d/a)=sqrt(2*...hm, wondering what the distance is along the lower block?
To find the acceleration of the 100 kg block during the time the 60 kg block remains in contact, we need to use Newton's second law of motion.
According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Therefore, we need to find the net force acting on the 100 kg block.
The only force acting on the 100 kg block is the force of friction between the two blocks. The force of friction can be calculated using the equation:
frictional force (Ff) = coefficient of friction (μ) * normal force (Fn)
Since there is no friction between the bottom 100 kg block and the horizontal frictionless surface, the normal force acting on the 100 kg block is equal to its weight (mass * gravitational acceleration). Hence, the frictional force between the two blocks is:
Ff = μ * m2 * g
where μ is the coefficient of friction between the two blocks (unknown), m2 is the mass of the 100 kg block, and g is the acceleration due to gravity.
Now, let's determine the coefficient of friction between the two blocks. We can use the definition of static friction to find this value.
The maximum static friction force (Fmax) is given by:
Fmax = μs * normal force (Fn)
In this case, the maximum static friction force is equal to the product of the mass of the lighter block (60 kg) and its acceleration (3.7 m/s^2):
Fmax = m1 * a1
Substituting the given values, we have:
Fmax = 60 kg * 3.7 m/s^2
Now, we can substitute this value for Fmax into the equation for frictional force (Ff):
Ff = μ * m2 * g
Since the blocks are in contact and moving together, the frictional force must be equal to the maximum static friction force (Fmax):
Ff = Fmax
μ * m2 * g = m1 * a1
Substituting the known values:
μ * 100 kg * 9.8 m/s^2 = 60 kg * 3.7 m/s^2
Now, solve for the coefficient of friction (μ):
μ = (60 kg * 3.7 m/s^2) / (100 kg * 9.8 m/s^2)
Calculating this, we find that μ is approximately 0.2265.
Finally, we can calculate the acceleration of the 100 kg block:
Ff = μ * m2 * g
a2 = Ff / m2
Substituting the known values:
a2 = (0.2265) * (100 kg * 9.8 m/s^2) / 100 kg
Calculating this, we find that the acceleration of the 100 kg block is approximately 2.213 m/s^2.