# Classical block on inclined plane

## Homework Statement

A block at t=0 is at the bottom of the plane and is projected up an inclined plane with initial speed v0. The plane's acute angle is θ above the horizontal, and the coefficient of friction is μ between the block and plane.

Find the time (t) and velocity (v1) of the block when it reaches a given distance d.

## Homework Equations

1. F=ma
2. d=(1/2)at^2 + v0t + d0
3. v1=at+v0

## The Attempt at a Solution

I drew a picture electronically: I made the direction FN and v0 positive.

#1
x dir
F = ma
F = -FμN - mgsinθ
ma = -μmgcosθ - mgsinθ
a = g(-μcosθ - sinθ)

So I found my a (should be a negative value), and now I want to find t when the block reaches d.

#2
d = (1/2)(-at^2) +v0t +d0
0 = (1/2)(-at^2) +v0t + 0 - d

Problem: Acceleration and distance are negative so the roots are imaginary, but I need to find real roots of t and then plug it in v1=at+v0 (I think).

Why do you have -a in equation #2?

And why do you think that the roots are imaginary? They depend on d; if it is sufficiently small, roots will be real. This makes perfect sense physically, because there is some maximum distance that the block can travel at any initial velocity.

haruspex
Homework Helper
Gold Member
Problem: Acceleration and distance are negative so the roots are imaginary, but I need to find real roots of t and then plug it in v1=at+v0 (I think).
That does not automatically make the roots imaginary. There will be real roots if v0 is nonzero and d is sufficiently small.

It's just that when I graph 0 = (1/2)(-at^2) +v0t + 0 - d, the entire curve is below the x-axis.

It's just that when I graph 0 = (1/2)(-at^2) +v0t + 0 - d, the entire curve is below the x-axis.

Consider ## a = 2, \ v_0 = 5, \ d = 4 ##: is your statement true?