Calculate acceleration & tension

[Melixa]

Homework Statement

A block with mass m1=15.0kg is on a rough inclined plane and is connected to an object with mass m2=25.0kg as shown. The rope may be considered massless; and the pulley may be considered frictionless. The coefficient of kinetic friction is µk=0.28

Homework Equations

a. what is the magnitude of the acceleration of the block?
b. what is the magnitude of the tension of the string?

The Attempt at a Solution

Attempt for question a: a=m2g-m1g sin theta-m1g cos theta/m1 + m2

Attempt for question b: T=W1+m1(a)

 

[haruspex]

Attempt for question a: a=m2g-m1g sin theta-m1g cos theta/m1 + m2

You've forgotten mu in there. It is in your attachment, but applied wrongly.

Attempt for question b: T=W1+m1(a)

What is W1? Anyway, your answer is wrong, both there and in the attachment. Please explain your reasoning.

 

[Saloni Khanna]

Since no perpendicular motion of mass m1 ,therefore equation :
R=m1g cos35 no applied.
Now, equation of motion of m1:
m1f=T-m1 g sin35 .
Equation of motion of m2:
m2f= m2g -T .
Adding the two equations,
(m1+m2)f=( m11 - m2sin35)*g
Rearranging,
f=m1-m2sin35*g divided by m1+m2.
Here f is acceleration.
Now tension T= m2(g-f)
Substituting f,
T= m1 m2(1+sin35)* g divided by m1+ m2.
Substituting ,m1= 15kg & m2= 25 kg ,sin35 = - 0.4281827 ,g=9.8 m/as.

Answer : f=6.297m/ss & T=52.53 m2f= m2g sin35degree.
Adding ,
(m1+m2)f=( m1- m2 sin35)g
Rearranging,
f=m1- m2sin35 *g divided by m1+m2.
Here f is acceleration.
Now tension T = m2(g-f)
Substituting f we get ,
T=m1*m2(1+sin35)*g divided by m1+m2.

 

[Melixa]

You've forgotten mu in there. It is in your attachment, but applied wrongly.

What is W1? Anyway, your answer is wrong, both there and in the attachment. Please explain your reasoning.

Sorry, we represent g with a W in class.

 

[Melixa]

Since no perpendicular motion of mass m1 ,therefore equation :
R=m1g cos35 no applied.
Now, equation of motion of m1:
m1f=T-m1 g sin35 .
Equation of motion of m2:
m2f= m2g -T .
Adding the two equations,
(m1+m2)f=( m11 - m2sin35)*g
Rearranging,
f=m1-m2sin35*g divided by m1+m2.
Here f is acceleration.
Now tension T= m2(g-f)
Substituting f,
T= m1 m2(1+sin35)* g divided by m1+ m2.
Substituting ,m1= 15kg & m2= 25 kg ,sin35 = - 0.4281827 ,g=9.8 m/as.

Answer : f=6.297m/ss & T=52.53m2f= m2g sin35degree.
Adding ,
(m1+m2)f=( m1- m2 sin35)g
Rearranging,
f=m1- m2sin35 *g divided by m1+m2.
Here f is acceleration.
Now tension T = m2(g-f)
Substituting f we get ,
T=m1*m2(1+sin35)*g divided by m1+m2.

So, I don't have to use the value of mu to calculate the acceleration and tension?

 

[haruspex]

Since no perpendicular motion of mass m1 ,therefore equation :
R=m1g cos35 no applied.
Now, equation of motion of m1:
m1f=T-m1 g sin35 .
Equation of motion of m2:
m2f= m2g -T .
Adding the two equations,
(m1+m2)f=( m11 - m2sin35)*g
Rearranging,
f=m1-m2sin35*g divided by m1+m2.
Here f is acceleration.
Now tension T= m2(g-f)
Substituting f,
T= m1 m2(1+sin35)* g divided by m1+ m2.
Substituting ,m1= 15kg & m2= 25 kg ,sin35 = - 0.4281827 ,g=9.8 m/as.

Answer : f=6.297m/ss & T=52.53
m2f= m2g sin35degree.
Adding ,
(m1+m2)f=( m1- m2 sin35)g
Rearranging,
f=m1- m2sin35 *g divided by m1+m2.
Here f is acceleration.
Now tension T = m2(g-f)
Substituting f we get ,
T=m1*m2(1+sin35)*g divided by m1+m2.

Saloni, you need to read the Homework Forum rules. We do not do students' homework for them. We provide hints, point out mistakes in working and in concepts, etc, and guide the student to the answer. the only time you should post a full solution is if the student already has a solution but you want to show a better one.
On this occasion, it happens that your solution is wrong. As Melixa notes, you forgot about friction. So not much harm done.

 

[haruspex]

Sorry, we represent g with a W in class.

Ok, but do you have any revisions to your answer in a? What is your reasoning for your answer in b?

 

[Saloni Khanna]

Sorry ,next I'll take care of it.
For part b) Tension is calculated from equation of motion of mass m2. N in part a) mu R=m1gcostheta.
Reply its correct answer pls. ma'am