Problem statement

Problem 1.62. A small body was launched up an inclined plane set at an angle α = 15° against the horizontal. Find the coefficient of friction, if the time of the ascent of the body is η = 2.0 times less than the time of its descent.

We assume g = 9.8 m/s² and a launch speed u = 5 m/s. The problem gives neither. They only set the real times and the distance, not k. Both are sliders.

We assume friction is sliding friction with one coefficient k, the same going up and coming down, and the body does not tip or roll.

Symbols: α is the angle of the plane, η is (time of descent)/(time of ascent), k is the friction coefficient, a₁ and a₂ are the accelerations going up and coming down, l is the distance climbed.

The big idea in plain words

Going up, gravity and friction both pull the body back down the slope, so it slows quickly. Coming down, gravity pulls it along but friction holds it back, so it speeds up slowly. The climb is therefore faster than the descent. The more friction there is, the bigger the gap between the two times.

The distance up and the distance down are the same. A body that covers the same distance with a bigger acceleration takes less time, and the time ratio squared equals the acceleration ratio. That single link turns a time ratio into a friction coefficient.

The solution in each frame

Only the ground frame is used. The ground is an inertial frame, so Newton's second law holds as written. A frame moving with the body would need a pseudo-force and give the same answer with more writing, so it adds nothing here.

The solution step by step

Step 1, going up. The normal force is m g cos α, so friction is k m g cos α. Both friction and the pull m g sin α point down the slope: a₁ = g (sin α + k cos α) = m/s². Using 0 = u² − 2 a₁ l, the distance is l = u²/(2 a₁) = m, reached after t₁ = u/a₁ = s.

Step 2, coming down. Friction now points up the slope: a₂ = g (sin α − k cos α) = m/s². Starting from rest, l = ½ a₂ t₂², so t₂ = s.

Step 3, compare. Going up, the body stops after l = ½ a₁ t₁² (the same motion reversed). So ½ a₁ t₁² = ½ a₂ t₂², and η² = (t₂/t₁)² = a₁/a₂ = (sin α + k cos α)/(sin α − k cos α) = .

Step 4, solve. η² sin α − η² k cos α = sin α + k cos α, so k cos α (η² + 1) = sin α (η² − 1). Then k = tan α (η² − 1)/(η² + 1) = × ( − 1)/( + 1) = .

Step 5, the possible cases. The body comes back only if k < tan α = . The formula always gives k below tan α, and k = 0 when η = 1. The return speed is a₂ t₂ = u/η = m/s.

Five methods

1. Force method

The solution above: two accelerations from Newton's second law, then the distance condition. It is the only method here that gives k directly.

2. Energy method

Going up, the body loses m u²/2 to gravity and friction: ½ m u² = (m g sin α + k m g cos α) l. Coming down it gains ½ m v² = (m g sin α − k m g cos α) l. Dividing gives (u/v)² = a₁/a₂ = η², so v = u/η. This gives the same ratio. It is Newton's law in energy form and needs the speeds, not the times.

3. Graph method

On the velocity graph the line goes down steeply to zero, then continues below zero with a gentle slope. The slope ratio is η², and the areas above and below the axis are equal, because the distances are equal. The page shows both.

4. Calculus method

Write x'' = −g (sin α + k cos α) going up and x'' = −g (sin α − k cos α) coming down. Integrate twice with x = 0 at the start and the turning point at x = l. The time ratio comes out as the square root of the acceleration ratio.

5. Check method

Units: k has no units, and tan α times a pure ratio has none. Limits: η = 1 gives k = 0, a smooth plane, equal times. As η grows, k tends to tan α, where friction just holds the body at the top. Putting k back gives a₁/a₂ = η², as the numerical check shows.

Fastest: the force method. Most reliable: the force method followed by the check method.

What each slider changes

Control	Changes the answer?	What it changes
Angle α	Yes	k through tan α, both accelerations, the distance, the times, the drawing and the graphs
Ratio η	Yes	k, both accelerations, the return speed and the graphs
Launch speed u	No	The real times, the distance climbed and the return speed. Not k
Gravity g	No	The accelerations, the real times and the distance. Not k
Given-answer mode	Sets η	The page finds the time ratio that goes with the chosen k
Time bar and Play	No	The clock in units of the ascent time, with 4 s holds at the top and at the return

Parameters

Quantity	Symbol	Value used
Angle of the plane	α	°
Time ratio	η
Launch speed	u	m/s
Gravity	g	m/s²
Friction coefficient	k
Acceleration up, down	a₁, a₂	, m/s²
Ascent time, descent time	t₁, t₂	s, s
Distance climbed	l	m
Return speed	u/η	m/s

Exam traps

• Using the same acceleration up and down. Friction reverses direction, so a₁ and a₂ differ.
• Writing t₂/t₁ = a₁/a₂ without the square root. Time depends on the square root of the ratio, so η² = a₁/a₂.
• Expecting the body to return with speed u. Friction takes energy both ways, so the return speed is u/η, smaller than u.
• Using α in radians in tan α on a calculator set to degrees. With 15° and η = 2, tan 15° = 0.268 and k = 0.16.
• Forgetting k must be below tan α. Above it the body would stay at the top and no descent time exists.