Two blocks of mass
m1 = 1.70 kg
and
m2 = 5.20 kg
are connected by a massless cord passing over a frictionless pulley as shown in the figure below, with
θ = 41.0°.
The coefficient of kinetic friction between block 1 and the horizontal surface is 0.400, and the coefficient of kinetic friction between block 2 and the inclined surface is 0.330.
(b) What is the acceleration of the system when it is released from rest?
(c) What is the tension in the cord connecting the blocks?
To find the acceleration of the system when it is released from rest and the tension in the cord connecting the blocks, we can use Newton's second law of motion and the concept of force and friction.
(b) To find the acceleration of the system, we need to consider the forces acting on the two blocks. We have one block (m1) on a horizontal surface and the other block (m2) on an inclined plane.
Start by analyzing the forces acting on m1:
1. The weight force (mg1) acting vertically downwards.
2. The normal force (N1) exerted by the surface in the opposite direction of weight force.
3. The friction force (f1) opposing the motion, which is equal to the coefficient of kinetic friction (μk1) multiplied by the normal force (f1 = μk1N1).
4. The tension force (T) acting on m1 and directed to the right due to cord.
For m2 on the inclined plane:
1. The weight force (mg2) acting vertically downwards.
2. The normal force (N2) exerted by the inclined plane in the perpendicular direction (opposite to the weight force).
3. The friction force (f2) opposing the motion, which is equal to the coefficient of kinetic friction (μk2) multiplied by the normal force (f2 = μk2N2).
4. The tension force (T) acting on m2 and directed upwards due to the cord.
Considering the forces along the x-direction, we can write the following equation using Newton's second law of motion:
(m2 * g2 - f2) - T = m2 * a
The force (m2 * g2 - f2) is the effective force moving m2 down the inclined plane. Now let's consider the forces along the y-direction:
N2 - m2 * g2 * cos(θ) = 0
Solving this equation for N2, we have:
N2 = m2 * g2 * cos(θ)
Using this value of N2, we can substitute it back into the equation for f2:
f2 = μk2 * N2 = μk2 * m2 * g2 * cos(θ)
Substituting these values back into the equation along the x-direction, we get:
(m2 * g2 - μk2 * m2 * g2 * cos(θ)) - T = m2 * a
Similarly, considering the forces along the y-direction for m1, we have:
N1 - m1 * g1 = 0
N1 = m1 * g1
Using this value of N1, we can substitute it into the equation for f1:
f1 = μk1 * N1 = μk1 * m1 * g1
The equation along the x-direction for m1 becomes:
T - μk1 * m1 * g1 = m1 * a
Now we have two equations for the two unknowns (T and a). Solve these equations simultaneously to find the values of T and a.
(c) To find the tension in the cord connecting the blocks, we can use the equation we derived earlier:
T = μk1 * m1 * g1 + m1 * a
Plug in the known values of μk1, m1, g1, and the value of a obtained from the previous calculation to find the tension in the cord.