Answer to Question #109142 in Mechanics | Relativity for Isaiah

Question

1) An object travels with a speed of 4 m/s toward a frictionless incline.
a) How high up the incline will the object travel?
b) If the angle of the incline is doubled (so the incline is steeper), how will this affect how high up the incline the object travels?
2) At the top of a 3 m high frictionless ramp, an object is moving at 2 m/s.
a) How fast will it be traveling when it reaches the bottom of the ramp?
b) How fast will the object be traveling when it is halfway down the ramp (i.e., at a height of 1.5 m above the ground)?

Answer 1

Explanation

• The law of "Conservation of mechanical energy" is applied throughout the question since the situation takes place within a friction less environment.

Notations

• Refer to the sketch attached.
• E_p & E_k: potential & kinetic energies respectively

Calculations

1.) Consider the sketch 1

a).Apply the law stated above from ground level to the height (h), the object will move.

$\qquad\qquad \begin{aligned} \small E_k +E_p&= \small E_{k1}+E_{p1}\\ \small \frac{1}{2}mv^2+0&=\small 0\,+ mgh\\ \small h&=\small \frac{v^2}{2g}&\cdots\cdots (\small1)\\ \small &= \small \frac{(4ms^{-1})^2}{2\times10}\\ \small &=\small \bold{0.8 m} \end{aligned}$

b). As the equation 1 describes, the height which the object travels depend only on the initial velocity & the g and not on the angle of inclination, meaning that doubling the inclination angle will not make any difference in the height which it may travel.

What happens is the distance that the object travels on the incline reduces.

$\qquad\qquad \begin{aligned} \small x&=\small \frac{h}{\sin(\theta)}\cdots\cdots( \small 1^{st} situation)\\ \small x_1&=\small \frac{h}{\sin(2\theta)} \cdots\cdots(2^{nd}situation)\\ \small \bold {x_1}&<\small \bold x \end{aligned}$

2).Consider the sketch 2

a).Apply the law stated above from top to bottom.

$\qquad\qquad \begin{aligned} \small \frac{1}{2}mv^2+mgh&=\small \frac{1}{2}mv_2^2\,+0\\ \small v_2^2&=\small \sqrt{2gh\,+v^2}\\ &= \small \sqrt{2\sdot10\sdot 3m\, +(2ms^{-1})^2}\\ &= \small \bold{8 ms^{-1}} \end{aligned}$

b).Apply the law from top to the halfway location

$\qquad\qquad \begin{aligned} \small \end{aligned}$ $\begin{aligned} \small \frac{1}{2}mv^2+mgh& = \small \frac{1}{2}mv_1^2+mgh\\ \small \frac{1}{2}(2)^2+10\sdot3&=\small \frac{1}{2}v_1^2+10\sdot(1.5)\\ \small \bold{v_1}&=\small \bold{5.83ms^{-1}} \end{aligned}$