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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
Question 2
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
Calculate the acceleration of the 20 kg mass.
Answer
To calculate the acceleration of the 20 kg mass, we first need to set up the equations of motion considering the forces acting on both masses.
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For the 5 kg mass (on the surface):
- The forces acting are:
- Tension (T) upwards
- Weight (mg) downwards, where ( m = 5 \text{ kg} )
- Frictional force (F_f) downwards = ( \mu_k \cdot N ), where ( N = mg )
- Equation: ( T - F_f - mg = 0 )
- This gives: ( T = \mu_k \cdot m , g + mg = (0.4)(5g) + 5g = 5(1 + 0.4)g = 7g )
- The forces acting are:
-
For the 20 kg mass (hanging):
- The forces acting are:
- Weight (mg) downwards
- Tension (T) upwards
- Equation: ( mg - T = ma )
- Rearranging gives: ( a = \frac{(20g - T)}{20} )
- Substituting for T: ( a = \frac{(20g - 7g)}{20} = \frac{13g}{20} \approx 6.47 \text{ m/s}^2 )
- The forces acting are:
Step 2
Calculate the speed of the 20 kg mass as it strikes the ground.
Answer
To find the speed of the 20 kg mass as it strikes the ground, we can use the kinematic equation:
[ v^2 = u^2 + 2as ]
Where:
- ( v ) is the final velocity,
- ( u ) is the initial velocity (0 m/s since it starts from rest),
- ( a ) is the acceleration (calculated as ( 6.47 \text{ m/s}^2 )),
- ( s ) is the distance (6 m from the initial height).
Substituting in:
[ v^2 = 0 + 2(6.47)(6) = 77.64 ]
Thus:
[ v \approx 8.81 \text{ m/s} ]
Step 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?
Answer
To find the minimal distance from the pulley, we need to consider the time it takes for the 20 kg mass to fall:
Using the kinematic equation: [ s = ut + \frac{1}{2}at^2 ] Where ( s = 6 \text{ m}, u = 0, a = 6.47 \text{ m/s}^2 ) gives:
[ 6 = 0 + \frac{1}{2}(6.47)t^2 ]
[ t^2 = \frac{12}{6.47} \approx 1.85 \Rightarrow t \approx 1.36 \text{ s} ]
Now, calculate the distance the 5 kg mass travels:
Using ( d = vt = 6.47 \cdot t ): [ d \approx 6.47 imes 1.36 \approx 8.80 \text{ m} ]
Thus, the 5 kg mass should be placed at least 8.80 m from the pulley.