Answer to Question #187199 in Physics for Nicole

Question

A 3.0 kg box is placed at the bottom of a frictionless ramp which is L = 2.5 m in length, angled at θ = 25° above horizontal. The box starts at rest and is pushed by a force FP =18.0 N  which acts parallel to the ramp. After the box leaves the top of the ramp, the force stops pushing and the box moves through the air as a projectile. 

a) Determine the acceleration of the box as it travels along the surface of the ramp.

b) Determine how long it takes the box to travel from the bottom of the ramp to the top.

c) Determine the speed of the box at the highest point of its projectile trajectory.

d) Determine how far the box moves horizontally (while in the air) before it returns to the height at which it left the top of the ramp.

Answer 1

(a) Let's apply the Newton's Second Law of Motion:

$$F_p-mgsin\theta=ma,$$ $$a=\dfrac{F_p-mgsin\theta}{m},$$ $$a=\dfrac{18.0\ N-3.0\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot sin25^{\circ}}{3.0\ kg}=1.86\ \dfrac{m}{s^2}.$$

(b) Let's first find the final speed of the box at the top of the ramp:

$$v^2=v_0^2+2ad,$$ $$v=\sqrt{v_0^2+2ad}=\sqrt{0+2\cdot1.86\ \dfrac{m}{s^2}\cdot2.5\ m}=3.05\ \dfrac{m}{s}.$$

Then, we can find the time that the box takes to travel from the bottom of the ramp to the top from the kinematic equation:

$$d=\dfrac{1}{2}(v_0+v)t,$$ $$t=\dfrac{2d}{v}=\dfrac{2\cdot2.5\ m}{3.05\ \dfrac{m}{s}}=1.64\ s.$$

(c) At the highest point of its projectile trajectory, there is only the horizontal component of the speed of the box:

$$v=v_x=v_0cos\theta=3.05\ \dfrac{m}{s}\cdot cos25^{\circ}=2.76\ \dfrac{m}{s}.$$

(d) Let's first find the height of the ramp from the geometry:

$$sin\theta=\dfrac{h}{d},$$ $$y_0=h=dsin\theta=2.5\ m\cdot sin25^{\circ}=1.06\ m.$$

We can find the time the box takes to return to the height at which it left the top of the ramp from the kinematic equation:

$$y=y_0+v_0tsin\theta-\dfrac{1}{2}gt^2,$$ $$1.06 =1.06+1.29t-4.9t^2,$$ $$1.29t-4.9t^2=0,$$ $$t=\dfrac{1.29\ \dfrac{m}{s}}{4.9\ \dfrac{m}{s^2}}=0.263\ s.$$

Finally, we can find the distance the box moves horizontally (while in the air) before it returns to the height at which it left the top of the ramp:

$$x=v_0tcos\theta,$$ $$x=3.05\ \dfrac{m}{s}\cdot 0.263\ s\cdot cos25^{\circ}=0.73\ m.$$