5. Blocks A and B are connected by a string passing over an essentially friction-
less pulley, as in Figure 6. When the blocks are in motion, Block A experiences
a force of kinetic friction of magnitude 5.7N. If mA 2.7kg and mB 3.7kg,
calculate the magnitude of
(a) the acceleration of the blocks
(b) the tension in the string
(a) To calculate the acceleration of the blocks, we first need to find the net force acting on each block.
For Block A:
- The force of gravity acting on Block A is given by Fg = mA * g = 2.7kg * 9.8m/s^2 = 26.46N
- The force of kinetic friction is given as fk = 5.7N
- The tension in the string is T
- The net force acting on Block A is given by: net force = T - fk - Fg
For Block B:
- The force of gravity acting on Block B is given by Fg = mB * g = 3.7kg * 9.8m/s^2 = 36.26N
- The tension in the string is T
- The net force acting on Block B is given by: net force = T - Fg
Since the two blocks are connected by a string over a pulley, the tension in the string is the same for both blocks. Therefore, we can write:
T - fk - Fg = mA * a
T - Fg = mB * a
Now, we can solve for acceleration a:
T - 5.7 - 26.46 = 2.7 * a
T - 36.26 = 3.7 * a
Adding the two equations, we get:
2T - 41.06 = 6.4a
2T = 6.4a + 41.06
Now, we substitute the value of T from the second equation into this equation:
2(3.7a + 36.26) = 6.4a + 41.06
7.4a + 72.52 = 6.4a + 41.06
1a = -31.46 / 1
a = -31.46m/s^2
Therefore, the acceleration of the blocks is 31.46m/s^2.
(b) To find the tension in the string, we can substitute the acceleration back into the second equation:
T - 36.26 = 3.7 * (-31.46)
T - 36.26 = -115.402
T = 115.402 - 36.26
T = 79.142N
Therefore, the tension in the string is 79.142N.