Answer:
d = 2.5 m
Explanation:
- The work-energy theorem, states that the change in the kinetic energy of a body, is equal to the work done on the object, by the net external force on it, as follows:
[tex]\Delta K = W_{net} =- (F_{net} * d_{rp}) (1)[/tex]
[tex]\Delta K = K_{f} - K_{0} = \frac{1}{2} * m * (v_{f} ^{2} - v_{0} ^{2} ) (2)[/tex]
- In this case, our givens are as follows:
- v₀ = 3.9 m/s vf = 0.48*3.9 m/s = 1.87 m/s Fnet = 0.24*m*9.8 m/s2.
- Replacing by the givens in (1) and (2), rearranging and simplifying common terms, we can solve for drp, as follows:
[tex]v_{f}^{2} -v_{0}^{2} = - 0.24*g* d_{rp} \\ \\ (1.87m/s)^{2} -(3.9m/s)^{2} = -0.24*9.8m/s2*d_{rp}[/tex]
[tex]d_{rp} =\frac{-11.7(m/s)2}{-2*0.24*9.8 m/s2} = 2.5 m[/tex]
