Answer to Question #224939 in Mechanics | Relativity for Nea

Question

A 18 kg block starts from rest and slides down a frictionless incline. When the block has slide a distance of 2.0 m , it’s speed is 1.8 m/s. at what angle above the horizontal is the inclined plane tilted

Answer 1

Explanations & Calculations

• The incline is frictionless so that no frictional force acts on it along the upward direction of the incline and the normal force also is acting perpendicular to the plane. So it is just a FREEFALL on the incline.
• We know the acceleration of verticle freefall is $\small g$ and what applies here is a resolution (the resolution of g parallel to the incline) of it: $\small g\sin\theta$, if the inclination is taken to be $\small \theta$.
• Apply $\small F= ma$ on the block for its motion down the incline, you will find the acceleration to be the same.
• Then, with this acceleration— a constant acceleration— it is a simple motion in which 4 of the motion equations could be applied.
• And selecting the most appropriate equation according to the data given & applying we can get the answer we want.
• Knowns, $\small m =18kg,\,u=0ms^{-1},\,v=1.8ms^{-1},\,a=g\sin\theta,\,\&\,s=2m.$
• Then applying $\small v^2=u^2 +2as$ to the motion down the incline,

$\qquad\qquad \begin{aligned} \small (1.8ms^{-1})^2&=\small (0ms^{-1})^2+2(g\sin\theta)(2m)\\ \small \sin\theta&=\small \frac{3.24}{4\times9.8ms^{-2}}\\ &=\small 0.0827\\ \small \theta &=\small \sin^{-1}(0.0827)\\ &=\small \bold{4.74^0} \end{aligned}$

• Energy conservation is another method you could use to get the answer.