Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. Assume the incline is frictionless and take m1 = 2.00 kg, m2 = 7.35 kg, and θ = 50.0°.

1) Find the magnitude of the accelerations of the objects. (m/s^2)

2)Find the tension in the string. (N)

3)Find the speed of each object 2.50 s after being released from rest. (m/s)

assume m1 moving down, so m2 is moving up and accelleration direction of each box is the same as its movement.

check forces act on m1
1. tension of the rope T up
2. m1g down
3. moving down with a
f = ma --> m1g - T = m1a ...(1)
or. T = m1g - m1a ... (1.1)

check forces act on m2
1. tension of the rope T up alone the incline. it must be equal to previous one because it is the same segment of the rope.
2. m2gsine(θ) down along the incline
3. moving up with a
f = ma --> T - m2gsine(θ) = m2a ...(2)

replace T from (1.1) to (2)
(m1g - m1a) - m2gsin(θ) = m2a
m1g - m2gsin(θ) = m2a + m1a
a = g(m1 - m2sin(θ))/(m1 + m2)

plug in value you will get the answer. if a is negative, it means direction on our assumption is not correct. it's the opposite (direction).

after get a, you will get T
and velocity use v = u + at

... i cannot post pict. draw the diagram will make you understand this better.

aristocles aristocles · Answer 2 · 2019-10-15T20:59:31-04:00

Answer:

Part a)

\(a = 3.81 m/s^2\)

Part b)

\(T = 27.2 N\)

Part c)

\(v_f = 9.53 m/s\)

Explanation:

For m2 mass along the inclined plane we can write

\(m_2g sin\theta - T = m_2 a\)

for m1 mass we can write equation in vertical direction

\(T - m_1 g = m_1 a\)

so we will have

\(m_2g sin\theta - m_1g = (m_1 + m_2) a\)

so we have

\(a = \frac{(m_2 sin\theta - m_1)g}{m_1 + m_2}\)

now plug in all data in it

\(a = \frac{(7.35 sin50 - 2)9.81}{7.35 + 2}\)

\(a = 3.81 m/s^2\)

Part b)

from above equation

\(T = m_1g + m_1 a\)

\(T = 2(9.81 + 3.81)\)

\(T = 27.2 N\)

Part c)

Speed of the object after t = 2.50 s

\(v_f = v_i + at\)

\(v_f = 0 + 3.81(2.50)\)

\(v_f = 9.53 m/s\)