A 6-kilogram mass is placed on a frictionless plane inclined 30° to the horizontal.
The mass is connected to another 10-kilogram mass by a light, inextensible string via a pulley, as shown in the diagram
The 6-kilogram mass is released from rest and slides up the plane.
Given the acceleration due to gravity is \(g\) m s\(^{-2}\), express the velocity, \(v\), of the 6 kilogram mass in terms of \(t\) as it moves up the plane. (3 marks)
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\(\text{Resolving forces on the 6 kg mass:}\)
\(\text{Let} \ \ F_d=\text{force down slope:}\)
| \(\sin 30^{\circ}\) | \(=\dfrac{F_d}{6 g}\) |
| \(F_d\) | \(=6 g\, \sin 30=3 g\) |
\(\text{Using} \ \ \sum F=m a\):
\(T-3 g=6 a\ \ldots\ (1)\)
\(\text{Consider 10 kg mass:}\)
\(10 g-T=10 a\ \ldots\ (2)\)
\(\text{Add (1) + (2):}\)
\(7 g=16 a \ \Rightarrow \ a=\dfrac{7}{16} g\)
\(v=\displaystyle \int \dfrac{7}{16}g \, dt=\dfrac{7}{16} g t+c\)
\(\text{When} \ \ t=0 \quad v=0 \ \Rightarrow \ c=0\)
\(\therefore v=\dfrac{7}{16}gt\)