5. Blocks A and B are connected by a string passing over an essentially friction-

Question

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

Answer 1

(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.