A block of mass 6kg slides on a horizontal plane against a constant resistance o

Question

A block of mass 6kg slides on a horizontal plane against a constant resistance of 12.6 N. A light inextensible string is attached to the block and, after passing over a smooth pulley, is attached a freely hanging sphere of mass 3kg. The part of the string between the block and the pulley is horizontal.
The tension in the string is T newton and the acceleration of the block and sphere is a ms-2.
Write down the equation of motion of the block and also the equation of motion of the sphere, each in terms of T and a?
Find the values of T and a?

Explanation / Answer

For the sphere: FNETs = 2kg (a m/s^2) downward = 2a N downward For the block: FNETb = T - 12.6 N towards the sphere/pulley Solve using the sphere: Assume g (acceleration due to gravity) to be 10 m/s^2 2(10) - T = FNETs 20 - T = 2a 2a + T = 20 FNETs = FNETb 2a = T - 12.6 System of equations: 2a - T = -12.6 2a + T = 20 4a = 7.4 a = 1.85 m/s^2 Substitue back into original equation: 2(1.85) + T = 20 3.7 + T = 20 T = 16.3 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.