Answer to Question #187180 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) Determine the acceleration of the box as it travels along the surface of the ramp:

$$a=\frac{F_\text{net}}m,\\\space\\ F_\text{net}=F-mg\sin\theta,\\\space\\ a=\frac{F-mg\sin\theta}m=\frac{18-3·9.8\sin25°}3=1.86\text{ m/s}^2.$$

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

$$t=\sqrt{\frac{2L}{a}}=\sqrt{\frac{2·2.5}{1.86}}=1.64\text{ s}.$$

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

$$u=v_x=v\cos\theta=at\cos\theta=2.76\text{ m/s}.$$

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. First, find the magnitude of the velocity vector directed 25° above the horizontal:

$$v=at.$$

Height of top of the ramp:

$$H=L\sin\theta.$$

Height above the top reached by the box:

$$h=\frac{v_y^2}{2g}=\frac{(v\sin\theta)^2}{2g}=\frac{(at\sin\theta)^2}{2g}.$$

Time to reach the highest point after the box left the ramp:

$$t_u=\frac{v_y}g=\frac{at\sin\theta}g.$$

Time to fall down from height $y=H+h$:

$$t_d=\sqrt{\frac{2y}g}=\sqrt{\frac{2[L\sin\theta+(at\sin\theta)^2/(2g)]}g}.$$

Total time of flight:

$$T=t_u+t_d=\frac{at\sin\theta}g+\sqrt{\frac{2[L\sin\theta+(at\sin\theta)^2/(2g)]}g},\\\space\\ T=0.614\text{ s}.$$

The horizontal distance traveled after the box left the ramp:

$$R=v_xT=atT\cos\theta =1.70\text{ m}.$$