Question

question 16
problem reference 6.3
a 6.00 - kg box slides from rest down a frictionless incline from a height of 5.00 m. a constant frictional force, introduced at point a, brings the block to rest at point b, 19.0 m to the right of point a.
what is the speed of the box just before reaching point a?
15.9 m/s
6.75 m/s
2.51 m/s
9.90 m/s

Explanation:

Step1: Apply conservation of mechanical energy

The initial mechanical energy \(E_{i}\) is potential energy \(E_{i}=mgh\) (since the box starts from rest, kinetic energy \(K_{i} = 0\)). The final mechanical energy \(E_{f}\) just before point \(A\) is kinetic energy \(E_{f}=\frac{1}{2}mv^{2}\) (assuming the reference level for potential energy is at point \(A\), so \(U_{f}=0\)). By the conservation of mechanical energy \(E_{i}=E_{f}\), we have \(mgh=\frac{1}{2}mv^{2}\).

Step2: Solve for \(v\)

Cancel out the mass \(m\) from both sides of the equation \(mgh=\frac{1}{2}mv^{2}\). We get \(gh = \frac{1}{2}v^{2}\). Then \(v=\sqrt{2gh}\). Substitute \(g = 9.8\space m/s^{2}\) and \(h=5.00\space m\) into the formula: \(v=\sqrt{2\times9.8\times5}\). Calculate \(2\times9.8\times5=98\), and \(\sqrt{98}\approx9.90\space m/s\).

Answer:

\(9.90\space m/s\)