Two blocks are connected by a (massless) string that passes over a (massless, frictionless) pulley as shown. Assume the interface between block A and the horizontal tabletop to be frictionless. If mA = 1.6 kg and mB = 1.9 kg, what is the tension in the string?Express your answer in N, to at least one digit after the decimal point.
Question
Two blocks are connected by a (massless) string that passes over a (massless, frictionless) pulley as shown. Assume the interface between block A and the horizontal tabletop to be frictionless. If mA = 1.6 kg and mB = 1.9 kg, what is the tension in the string?Express your answer in N, to at least one digit after the decimal point.
Solution
The problem seems to be missing a diagram, but I'll assume a common setup where block A is on a horizontal table and block B is hanging off the edge of the table, connected by the string over the pulley.
The tension in the string can be found by considering the forces acting on each block.
For block B (the hanging block), the forces are its weight (mBg, where g is the acceleration due to gravity) and the tension in the string (T). Since block B is accelerating downwards, the net force on it is mBg - T.
For block A (the block on the table), the only horizontal force is the tension in the string, so the net force on it is T.
Since the blocks are connected by a string, they have the same magnitude of acceleration (a). Therefore, we can set up the following equations based on Newton's second law (F = ma):
For block A: T = mAa (1) For block B: mBg - T = mB*a (2)
We can solve these two equations simultaneously to find the tension T.
First, solve equation (1) for a: a = T/mA.
Then, substitute this into equation (2) to get: mBg - T = mB(T/mA).
Solving this equation for T gives: T = mB*g/(1 + mB/mA).
Substituting the given values (mA = 1.6 kg, mB = 1.9 kg, g = 9.8 m/s^2) gives: T = 1.9*9.8/(1 + 1.9/1.6) = 11.7 N.
So, the tension in the string is approximately 11.7 N.
Similar Questions
A block of mass 10 kg is suspended by an inextensible string passing over a smooth frictionless pulley. If the mass is pulled up at an acceleration of 1 m/s², calculate the tension in the string.
A mass m1 is connected by a light string that passes over a pulley of mass M to a mass m2 sliding on a frictionless horizontal surface as shown in the figure. There is no slippage between the string and the pulley. The pulley has a radius of 25.0 cm and a moment of inertia of ½ MR2. If m1 is 4.00 kg, m2 is 4.00 kg, and M is 4.00 kg, then what is the tension in the string attached to m2?
Two wooden blocks of masses 5 kg and 3 kg are connected by a massless rope. The 5 kg block is pulled by a horizontal force of 40 N.[All floors are frictionless. Ignore the vertical forces.](i) Find the acceleration of the blocks.(ii) Find the tension in the string
Two masses m1 and m2 are attached to a string which passes over a frictionless smooth pulley. When m1 = 10 kg, m2 = 6 kg, the acceleration of masses is gn𝑔𝑛. Value of n, is
Two masses, m1 and m2, are connected to each other as shown above. Mass m1 slides without friction on the table surface. Both masses have acceleration of magnitude a as shown.How does the tension in the string compare to the weight, m2g, of mass m2?Group of answer choicesIt depends on m1 being smaller than m2.The tension is equal to m2g.It depends on m1 being larger than m2.The tension is larger than m2g.The tension is smaller than m2g.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.