Question 5.4: Sam and his sister are sledging, but Sam wants to ride by hi...
Sam and his sister are sledging, but Sam wants to ride by himself. His sister gives him a push at the top of a smooth straight 15° slope and lets go when he is moving at 2 ms^{-1}. He continues to slide for 5 seconds before using his feet to produce a braking force of 95 N parallel to the slope. This brings him to rest. Sam and his sledge have a mass of 30 kg.
How far does he travel altogether?
To answer this question, you need to know Sam’s acceleration for the two parts of his journey. These are constant so you can then use the constant acceleration formulae.
Sliding freely
Using Newton’s second law in the direction of the acceleration gives
\begin{matrix}30g sin 15° = 30a_{1} &\longleftarrow \boxed{\text{Resultant force down the plane = mass × acceleration.}}\end{matrix}
a_{1} = 2.54
Now you know a_{1} you can find how far Sam slides (s_{1} m) and his speed (v ms^{-1}) before braking.
s = ut + \frac{1}{2}at²
\begin{matrix} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{\boxed{\text{Given u = 2, t = 5, a = 2.54}}}{\downarrow }\\ s_{1}= 2 × 5 + 1.27 × {25} = 41.75\end{matrix}
v = u + at
v = 2 + 2.54 × 5 = 14.7
Braking
By Newton’s second law down the plane
Resultant force = mass × acceleration
30g sin 15° – 95 = 30a_{2}
a_{2} = -0.63
v² = u² + 2as
\begin{matrix} 0 = 14.7² – 2 \times 0.63 \times s_{2} \longleftarrow & \boxed{\text{Given } u = 14.7, v = 0,}\end{matrix}
s_{2} = \frac{14.7²}{1.26} = 171.5
Sam travels a total distance of 41.75 + 171.5 m = 213 m to the nearest metre.
