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. The diagram is not drawn to scale. 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?

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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 - NSC Physical Sciences - Question 2 - 2016 - Paper 1

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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, we analyze the forces acting on both masses. Starting with the 20 kg mass:

The force of gravity acting on the 20 kg mass is: $F_{g_{20}} = m_{20} imes g = 20 imes 9.8 = 196 \, \text{N}$

The force of tension (T) exerted by the string is also acting upward. For the 5 kg mass, the forces are:

Weight down: $F_{g_{5}} = m_{5} \times g = 5 \times 9.8 = 49 \, \text{N}$
Friction force opposing motion: $F_{friction} = \mu_k \times F_{normal} = 0.4 \times (5 \times 9.8) = 19.6 \, \text{N}$

Now, setting up the equation for the forces: $T - F_{friction} = m_{5} a$ $196 - T = m_{20} a$

By substituting and solving these equations, we have:

$T = m_{5} g - F_{friction}$
Substituting into the second equation allows us to express it in terms of acceleration (a): $196 - (5 \times 9.8 - 19.6) = 20a$

By simplifying: $176.4 = 20a \Rightarrow a = \frac{176.4}{20} = 8.82 \, \text{m/s}^2$

Step 2

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

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Answer

We can use the kinematic equations to find the final velocity ( $v_f$ ) just before the mass strikes the ground. The equation is: $v_f^2 = u^2 + 2as$ where:

$u$ is the initial velocity (0 m/s),
$a$ is the acceleration (8.82 m/s²),
$s$ is the distance fallen (6 m).

Substituting the known values: $v_f^2 = 0 + 2(8.82)(6) \Rightarrow v_f = \sqrt{105.84} = 10.29 \, \text{m/s}$

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

To ensure the 20 kg mass just strikes the ground, we need to calculate the time (t) it takes for the 20 kg mass to fall the 6 m: $s = ut + \frac{1}{2} at^2$ Given that $u = 0$ : $6 = \frac{1}{2}(8.82)t^2 \Rightarrow t^2 = \frac{12}{8.82} \Rightarrow t \approx 1.23 \, \text{s}$

During this time, the 5 kg mass will also move, and its distance covered (d) can be calculated using the equation: $d = ut + \frac{1}{2} at^2 = 0 + \frac{1}{2}(8.82)(1.23^2) \approx 5.3 \, \text{m}$

Thus, the 5 kg mass should be placed at a minimum of approximately 5.3 m from the pulley.

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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