I cant solve this problem, and our teacher said that this would be in the test we’ll have tomorrow, can someone help me?
A body of m = 6.8kg is launched with a speed of 7.5 m / s towards the top of an inclined plane of 15 ° with respect to the horizontal. in the absence of friction, what displacement does it make before reversing the direction of motion?
Question
Answer:
d = 11.1 m
Explanation:
Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:
[tex] \frac{1}{2} m {v}^{2} = mgh[/tex]
Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:
[tex] \sin(15) = \frac{h}{d} \\ or \: h = d \sin(15) [/tex]
Plugging this into the energy conservation equation and cancelling m, we get
[tex] {v}^{2} = 2gd \sin(15)[/tex]
Solving for d,
[tex]d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m[/tex]