Question

A 5.0 kg mass moves along a horizontal surface when connected by a string passing over a pulley to a 2.5 kg mass. If its coefficient of kinetic friction is 0.25, what is its acceleration and the tension in the string?

Answer 1

Answer on Question 68545, Physics, Other

Question:

A $5.0 \, kg$ mass moves along a horizontal surface when connected by a string passing over a pulley to a $2.5 \, kg$ mass. If its coefficient of kinetic friction is 0.25, what is its acceleration and the tension in the string?

Solution:

a) Let $m_A = 5.0 \, kg$ and $m_B = 2.5 \, kg$ . Applying the Newton Second Law of Motion we get:

$$\sum F _ {x} = m _ {A} a _ {x},$$ $$T - F _ {f r} = m _ {A} a,$$ $$T - \mu_ {k} m _ {A} g = m _ {A} a (1).$$ $$\sum F _ {y} = m _ {B} a _ {y},$$ $$m _ {B} g - T = m _ {B} a. (2)$$

Let's express $T$ from the equation (1) and substitute it into the equation (2):

$$T = \mu_ {k} m _ {A} g + m _ {A} a,$$ $$m _ {B} g - \mu_ {k} m _ {A} g - m _ {A} a = m _ {B} a,$$ $$m _ {B} g - \mu_ {k} m _ {A} g = (m _ {A} + m _ {B}) a,$$ $$a = \frac {g (m _ {B} - \mu_ {k} m _ {A})}{m _ {A} + m _ {B}} = \frac {9 . 8 \frac {m}{s ^ {2}} \cdot (2 . 5 k g - 0 . 2 5 \cdot 5 . 0 k g)}{5 . 0 k g + 2 . 5 k g} = 1. 6 3 \frac {m}{s ^ {2}}.$$

b) Finally, substituting $a$ into the equation for $T$ we get:

$$T = m _ {A} (\mu_ {k} g + a) = 5. 0 k g \cdot \left(0. 2 5 \cdot 9. 8 \frac {m}{s ^ {2}} + 1. 6 3 \frac {m}{s ^ {2}}\right) = 2 0. 4 N.$$

Answer:

a) $a = 1.63\frac{m}{s^2}$

b) $T = 20.4N$

Answer provided by https://www.AssignmentExpert.com