3Thanks
• 2 Post By ebaines
• 1 Post By skeeter

# Thread:

1. ## Mechanics forces acting on a object in a slope

answer to q15 is 2.42ms can anyone solve these questions please

2. ## Re: Mechanics forces acting on a object in a slope

Originally Posted by abdulrehmanshah
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

3. ## Re: Mechanics forces acting on a object in a slope

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

4. ## Re: Mechanics forces acting on a object in a slope

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.

5. ## Re: Mechanics forces acting on a object in a slope

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.

6. ## Re: Mechanics forces acting on a object in a slope

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: $\small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)$

7. ## Re: Mechanics forces acting on a object in a slope

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: $\small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)$

8. ## Re: Mechanics forces acting on a object in a slope

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: $\small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)$

9. ## Re: Mechanics forces acting on a object in a slope

"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: $\frac 1 2 mv_2^2 - \frac 1 2 mv_1^2$.

10. ## Re: Mechanics forces acting on a object in a slope

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 ...

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

$v_f = 2.42 \, m/s$

11. ## Re: Mechanics forces acting on a object in a slope

how to change the topic to solved
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