Question

Objects of masses m1=4.00kg and m2 = 9.00kg are connected by a light string that passes over a frictionless pulley. The object m1 is held at rest on the floor, and m2 rests on a fixed incline of (theta)=40.0 degree. The objects are released from rest, and m2 slides 1.00m down the incline in 4.00s. Determine

a) the acceleration of each object,

b) the tension in the string, and

c) the coefficient of kinetic friction between m2 and he incline.

Answer 1

The acceleration of each object is $0.125m/s^2$

Further explanation

Acceleration is velocity rate of change of of an object with respect to time. An object's acceleration is the net result of all forces acting on the object, as described by Newton's Second Law. Whereas kinetic friction is a force that acts between moving surfaces. Then tension is the pulling force transmitted axially by the means of a string, a cable, chain etc

Objects of masses $m_1=4.00kg$ and $m_2 = 9.00kg$ are connected by a light string that passes over a frictionless pulley. The object $m_1$ is held at rest on the floor, and $m_2$ rests on a fixed incline of $(\theta)=40.0 degree$. The objects are released from rest, and $m_2$ slides $1.00m$ down the incline in $4.00s.$ Determine

a) the acceleration of each object,

$a= \frac{2x}{(t)^2} \\ a=\frac{2*1}{(4)^2}  =0.125m/s^2$

b) the tension in the string, and

$T - m_1 g = m_1 a \\T = m_1(g + a) \\= 4.00(9.8 + 0.125) \\= 39.7 N.$

c) the coefficient of kinetic friction between $m_2$ and he incline.

$m_2 *g* sin(\theta) - T - uR = m_2 a \\m_2 *g* cos(\theta) = R$

$m_2 *g(sin(\theta) - u cos(\theta)) - T = m_2 a \\m_2 g sin(\theta) - m2 a - T = m_2 g*u* cos(\theta)\\u = \frac{m_2(g sin(\theta) - a) - T}{m_2 g cos(\theta)}  \\u=\frac{9.00(9.81 sin(40) - 0.125) - 39.7}{9.00 * 9.81 cos(40)} \\u=   \frac{15.926}{67.634} = 0.235$

Learn more

1. Learn more about kinetic friction https://brainly.com/question/12881684
2. Learn more about acceleration https://brainly.com/question/11202599
3. Learn more about tension https://brainly.com/question/2386400

Answer details

Grade:  9

Subject:    physics

Chapter:  acceleration

Keywords: object,  acceleration, tension,  kinetic friction, slides

Answer 2

The acceleration of each object is equal to 0.125m/². The tension in the string is 39.7N and the coefficient of kinetic friction is equal to 0.235.

We can arrive at this answer as follows:

• To calculate the acceleration of the object we will use the following formula:

$a=\frac{2x}{(t)^2} \\a= \frac{2*1}{(4)^2} \\a= 0.125m/s^2$

• To calculate the tension in the string we will use the following formula:

$T= m1*(g+a)\\T= 4.00*(9.8+0.125)\\T= 39.7N$

• With these two values, we can calculate the coefficient of kinetic friction between m2 and the incline. For this, we will use the following formula:

$u= \frac{m2*[g*sin(theta)-a]- T}{m2*g*cos(theta)} \\u=\frac{15.926}{67.634} \\u= 0.235$

More information:

https://brainly.com/question/11348644?referrer=searchResults