(a) An object is thrown up a smooth plane inclined at an angle of 30° to the horizontal. If the plane is 15m long and the object comes to rest at the top, f...

Answer Details

(a) Object on a smooth inclined plane (angle \(30^\circ\), length \(15\ \text{m}\), comes to rest at the top). On a smooth plane the only force along the plane is the component of gravity, giving a retardation

\[a=g\sin30^\circ=10\times0.5=5\ \text{m s}^{-2}\quad(\text{taking }g=10\ \text{m s}^{-2}).\]

(i) Initial speed. Using \(v^2=u^2-2as\) with \(v=0\), \(s=15\ \text{m}\):

\[0=u^2-2(5)(15)\;\Rightarrow\;u^2=150\;\Rightarrow\;u=\sqrt{150}\approx 12.25\ \text{m s}^{-1}.\]

(ii) Time to reach the top. Using \(v=u-at\) with \(v=0\):

\[0=12.25-5t\;\Rightarrow\;t=\frac{12.25}{5}\approx 2.45\ \text{s}.\]

(b) Five forces on body P. From the diagram the forces are symmetric about the horizontal \(x\)-axis. Measuring angles from the positive \(x\)-axis (each gap is \(30^\circ\)):

• \(5\ \text{N}\) at \(60^\circ\)
• \(5\sqrt3\ \text{N}\) at \(30^\circ\)
• \(10\ \text{N}\) at \(0^\circ\)
• \(5\sqrt3\ \text{N}\) at \(-30^\circ\)
• \(5\ \text{N}\) at \(-60^\circ\)

(i) Magnitude of the resultant. Resolve horizontally:

\[\sum F_x = 5\cos60^\circ+5\sqrt3\cos30^\circ+10+5\sqrt3\cos30^\circ+5\cos60^\circ\]

\[\sum F_x = 2.5+7.5+10+7.5+2.5 = 30\ \text{N}.\]

Vertically the pairs cancel by symmetry:

\[\sum F_y = 5\sin60^\circ+5\sqrt3\sin30^\circ-5\sqrt3\sin30^\circ-5\sin60^\circ = 0.\]

Hence the resultant is

\[R=\sqrt{30^2+0^2}=\mathbf{30\ \text{N}}\ \text{along the }10\text{N direction (positive }x\text{-axis).}\]

(ii) Acceleration of the body (mass \(5\ \text{kg}\)):

\[a=\frac{R}{m}=\frac{30}{5}=\mathbf{6\ \text{m s}^{-2}}\ \text{in the direction of the }10\text{N force.}\]