A 5 kg mass and a 20 kg mass are connected by a light inextensible string which passes over a light frictionless pulley. Initially, the 5 kg mass is held stationary on a horizontal surface, while the 20 kg mass hangs vertically downwards, 6 m above the ground, as shown in the diagram below. When the stationary 5 kg mass is released, the two masses begin to move. The coefficient of kinetic friction, $\\mu_k$, between the 5 kg mass and the horizontal surface is 0.4. Ignore the effects of air friction. 2.1.1 Calculate the acceleration of the 20 kg mass. 2.1.2 Calculate the speed of the 20 kg mass as it strikes the ground. 2.1.3 At what minimum distance from the pulley should the 5 kg mass be placed initially, so that the 20 kg mass just strikes the ground?

NSC Physical Sciences Mechanics: Force and Newton’s Laws: A 5 kg mass and a 20 kg mass are

A 5 kg mass and a 20 kg mass are connected by a light inextensible string which passes over a light frictionless pulley - NSC Physical Sciences - Question 2 - 2016 - Paper 1

Question 2

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A 5 kg mass and a 20 kg mass are connected by a light inextensible string which passes over a light frictionless pulley. Initially, the 5 kg mass is held stationary ... show full transcript

Worked Solution & Example Answer:A 5 kg mass and a 20 kg mass are connected by a light inextensible string which passes over a light frictionless pulley - NSC Physical Sciences - Question 2 - 2016 - Paper 1

Step 1

2.1.1 Calculate the acceleration of the 20 kg mass.

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Answer

To calculate the acceleration of the 20 kg mass, we analyze the forces acting on both masses. For the 5 kg mass on the horizontal surface, we can write the equation:

$T - \\mu_k m_1 g = m_1 a$

For the 20 kg mass:

$m_2 g - T = m_2 a$

Substituting the known values:

(20)(9.8) - T = 20a$$ Solving these equations simultaneously, we substitute T from the first into the second. We find: 1. From the first equation, derive T: $$T = 5a + 0.4(5)(9.8)$$ 2. Substitute T into the second equation: $$(20)(9.8) - (5a + 0.4(5)(9.8)) = 20a$$ 3. Solve for a: $$a = 7.06 \\text{ m/s}^2$$

Step 2

2.1.2 Calculate the speed of the 20 kg mass as it strikes the ground.

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Answer

Using the kinematic equation:

$v^2 = u^2 + 2as$

Where:

$v$ is the final speed,
$u$ is the initial speed (0 m/s),
$a$ is the previously calculated acceleration (7.06 m/s²),
$s$ is the distance fallen (6 m):

Substituting the values, we get:

$v^2 = 0 + 2(7.06)(6) = 84.72 \\Rightarrow v = \sqrt{84.72} = 9.2 \\text{ m/s}$

Step 3

2.1.3 At what minimum distance from the pulley should the 5 kg mass be placed initially, so that the 20 kg mass just strikes the ground?

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Answer

The distance can be calculated using the relation:

$d = rac{1}{2} a t^2$

First, we need to determine time (t) taken for the 20 kg mass to reach the ground: Using: $v = u + at$

And from the previous step, we have: $9.2 = 0 + (7.06)t \\Rightarrow t = \frac{9.2}{7.06} = 1.3 \\text{ s}$

Now substituting t back to calculate the distance:

$d = rac{1}{2} (7.06)(1.3)^2 = 6 m$

Thus, the minimum distance the 5 kg mass should be placed initially is equal to the height it will fall, which is 6 m.

A 5 kg mass and a 20 kg mass are connected by a light inextensible string which passes over a light frictionless pulley - NSC Physical Sciences - Question 2 - 2016 - Paper 1

Worked Solution & Example Answer:A 5 kg mass and a 20 kg mass are connected by a light inextensible string which passes over a light frictionless pulley - NSC Physical Sciences - Question 2 - 2016 - Paper 1

2.1.1 Calculate the acceleration of the 20 kg mass.

2.1.2 Calculate the speed of the 20 kg mass as it strikes the ground.

2.1.3 At what minimum distance from the pulley should the 5 kg mass be placed initially, so that the 20 kg mass just strikes the ground?

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