Physics, 09.03.2020 01:33 anthonyjackson12aj

Could you help me with the question in the photo attached please.

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Answer from: crookdamian21

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

b) Yes

c) $a=\frac{m_2 g-m_1 g sin \theta - \mu_k m_1 g cos \theta}{m_1+m_2}$

Explanation:

To find the expression for the acceleration of the blocks, we have to write the equations of the forces acting on the two blocks.

In the following, we assume that the system accelerates downward (block 2) and up along the ramp (block 1).

For block 1, the equation of motion is:

$T-m_1 g sin \theta = m_1 a$

where:

T is the tension in the string (up along the ramp)

$m_1 g sin \theta$ is the component of the weight parallel to the ramp (down along the ramp), with

m_1 =mass of the block

= acceleration due to gravity

$\theta$ = angle of the ramp

is the acceleration of the system

For block 2, we have:

m_2 g-T = m_2 a

where

m_2 g is the weight of block 2, where

m_2 is the mass of block 2

From eq. (2) we get

T=m_2 g- m_2 a

And substituting into eq.(1), we find an expression for the acceleration of the system:

$m_2 g- m_2 a -m_1 g sin \theta = m_1 a\\(m_1+m_2)a=m_2g-m_1 g sin \theta\\a=\frac{m_2 g-m_1 g sin \theta}{m_1+m_2}$

We could have guessed the expression for the acceleration using Newton's second law:

F=ma

where, in this case:

F is the net force on the system, which is the difference between the weight of the block 2 and the component of the weight of block 1 acting along the ramp, so

$F=m_2 g- m_1 g sin \theta$

m is the total mass of the system, which is the sum of the masses of the two blocks:

m=m_1 + m_2

a is the acceleration

And solving for a:

$a=\frac{F}{m}=\frac{m_2 g-m_1 g sin \theta}{m_1 + m_2}$

In this case, there is also kinetic friction acting along the ramp, on block 1.

The magnitude of the kinetic friction is:

$F_f=\mu_k N$

where

$\mu_k$ is the coefficient of kinetic friction

N is the normal force acting on block 1

The normal force on block 1 can be found by writing the equation of the forces on the direction perpendicular to the ramp, we have:

$m_1g cos \theta - N = 0$

where $m_1g cos \theta$ is the component of the weight of the block perpendicular to the ramp. Therefore,

$N=m_1 g cos \theta$

So the force of friction is

$F_f=\mu_k m_1 g cos \theta$

The direction of this force is in the direction opposite to the motion (up along the ramp), so the equation of motion for block 1 becomes

$T-m_1 g sin \theta - \mu_k m_1 g cos \theta = m_1 a$

Substituting again the expression for the tension obtained in part a), we get:

$m_2 g-m_2 a-m_1 g sin \theta - \mu_k m_1 g cos \theta = m_1 a\\a=\frac{m_2 g-m_1 g sin \theta - \mu_k m_1 g cos \theta}{m_1+m_2}$

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