Question

Pushing one body toward the inclined plane, which formed the angle α, at speed v. The body
Having reached its climax, it returned to its original position. Body and inclined plane
coefficient of friction between k. When the body returns to its original position
at what speed? What time will it take?

Answer 1

$$0.5mv^2=mgl\sin \alpha+\mu mgl$$

$$l=\frac{v^2}{2g(\mu+\sin \alpha)}$$

Then,

$$0.5mu^2=0.5mv^2-2\mu mgl$$

$$u^2=v^2-4\mu g\frac{v^2}{2g(\mu+\sin \alpha)}=v^2\frac{1-2\mu}{(\mu+\sin \alpha)}$$

$$u=v\sqrt{\frac{1-2\mu}{(\mu+\sin \alpha)}}$$

1)

$$t=t_1+t_2=\frac{v}{g(\sin \alpha+\mu)}+\frac{u}{g(\sin \alpha-\mu)}$$

$$t=\frac{v}{g(\sin \alpha+\mu)}+\frac{v}{g(\sin \alpha-\mu)}\sqrt{\frac{1-2\mu}{(\mu+\sin \alpha)}}$$