Mechanics forces acting on a object in a slope

**abdulrehmanshah** · November 4th 2012, 06:25 PM

$Mechanics forces acting on a object in a slope-123.jpg$ answer to q15 is 2.42ms can anyone solve these questions please

**topsquark** · November 4th 2012, 06:38 PM

Originally Posted by abdulrehmanshah

$Click image for larger version. Name: 123.JPG Views: 13 Size: 54.8 KB ID: 25548$ answer to q15 is 2.42ms can anyone solve these questions please

First, one question per thread.

Second, these are all Newton's 2nd Law problems. Can you show us how you set this up?

-Dan

**abdulrehmanshah** · November 5th 2012, 07:12 AM

well can u please solve question 15 only once u solve it i will make change the topic to solved ok i copied it from a book

Originally Posted by topsquark

First, one question per thread.

Second, these are all Newton's 2nd Law problems. Can you show us how you set this up?

-Dan

**skeeter** · November 5th 2012, 08:06 AM

forces acting on mass A ...

$T - Mg\sin{\theta} = Ma$

forces acting on mass B ...

$mg - T = ma$

combining equations ...

$mg - Mg\sin{\theta} = Ma + ma$

$\frac{m - M\sin{\theta}}{M + m} \cdot g = a$

this acceleration will be constant as mass A slides up 2.8 m ... use an appropriate kinematics equation to find the velocity of mass A when mass B hits the ground.

for the last 0.2 m, the only force acting on mass A will be the component of weight acting parallel to the incline, so mass A's acceleration will change ...

$g\sin{\theta} = a_2$

use the velocity found when mass B hits the ground, the acceleration $a_2$ , and the remaining displacement to find the final velocity of mass A when it hits the pulley.

**abdulrehmanshah** · November 5th 2012, 08:30 AM

this is where i got to but i cant get to the final velocity as 2.42ms^-1 can u please help me here
i write equation 2as=v^2-u^2.
2*4.9*0.2=v^2-u^2
what should be u(i.e initial velocity)

Originally Posted by skeeter

forces acting on mass A ...

$T - Mg\sin{\theta} = Ma$

forces acting on mass B ...

$mg - T = ma$

combining equations ...

$mg - Mg\sin{\theta} = Ma + ma$

$\frac{m - M\sin{\theta}}{M + m} \cdot g = a$

this acceleration will be constant as mass A slides up 2.8 m ... use an appropriate kinematics equation to find the velocity of mass A when mass B hits the ground.

for the last 0.2 m, the only force acting on mass A will be the component of weight acting parallel to the incline, so mass A's acceleration will change ...

$g\sin{\theta} = a_2$

use the velocity found when mass B hits the ground, the acceleration $a_2$ , and the remaining displacement to find the final velocity of mass A when it hits the pulley.

**ebaines** · November 5th 2012, 08:41 AM

I suggest that you consider conservation of energy principles. There are two phases to this problem - the first phases is the 4 Kg mass being pulled up the ramp by the force of the 3Kg mass, and then the second phase is after the 3Kg mass hits the ground the tension in the rope becomes zero and the 4Kg mass continues to coast up uphill but slows as it rises. For phase 1 you have $\small\Delta KE + \Delta PE = 0$ , where $\small \Delta KE = \frac 1 2 (m_A + M_B) v_1^2$ and $\small \Delta PE = m_A g (2.8m) \sin 30 = m_B g 2.8m$ , so you can solve for $v_1$ . For phase 2: $\small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)$

**abdulrehmanshah** · November 5th 2012, 08:48 AM

what is the value of m

Originally Posted by ebaines

I suggest that you consider conservation of energy principles. There are two phases to this problem - the first phases is the 4 Kg mass being pulled up the ramp by the force of the 3Kg mass, and then the second phase is after the 3Kg mass hits the ground the tension in the rope becomes zero and the 4Kg mass continues to coast up uphill but slows as it rises. For phase 1 you have $\small\Delta KE + \Delta PE = 0$ , where $\small \Delta KE = \frac 1 2 (m_A + M_B) v_1^2$ and $\small \Delta PE = m_A g (2.8m) \sin 30 = m_B g 2.8m$ , so you can solve for $v_1$ . For phase 2: $\small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)$

**abdulrehmanshah** · November 5th 2012, 09:12 AM

K.E=0.5*m*V where V is final velocity what have u done with phase 2

Originally Posted by ebaines

I suggest that you consider conservation of energy principles. There are two phases to this problem - the first phases is the 4 Kg mass being pulled up the ramp by the force of the 3Kg mass, and then the second phase is after the 3Kg mass hits the ground the tension in the rope becomes zero and the 4Kg mass continues to coast up uphill but slows as it rises. For phase 1 you have $\small\Delta KE + \Delta PE = 0$ , where $\small \Delta KE = \frac 1 2 (m_A + M_B) v_1^2$ and $\small \Delta PE = m_A g (2.8m) \sin 30 = m_B g 2.8m$ , so you can solve for $v_1$ . For phase 2: $\small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)$

**ebaines** · November 5th 2012, 09:58 AM

"m" stands for "mass." You are told that the mass of block A is 4 Kg and the mass of block B is 3 Kg.

The change in kinetic energy of an object is: $\frac 1 2 mv_2^2 - \frac 1 2 mv_1^2$ .

**skeeter** · November 5th 2012, 01:02 PM

for the initial motion 2.8 m up the incline ...

$v_f^2 = v_0^2 + 2a(\Delta x)$

$v_f = \sqrt{2a(\Delta x)} = \sqrt{\frac{2g(2.8)}{7}} = \, 2.8 m/s$

for the final 0.2 m up the incline ...

$v_f^2 = 2.8^2 - g(0.2)$

$v_f = 2.42 \, m/s$

**abdulrehmanshah** · November 5th 2012, 07:59 PM

how to change the topic to solved

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