Question

The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree incline and achieve a speed of 62 km/h at the end of 75 m

Answer 1

Hi there!

We must begin by converting km/h to m/s using dimensional analysis:

$\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s$

Now, we can use the kinematic equation below to find the required acceleration:

vf² = vi² + 2ad

We can assume the object starts from rest, so:

vf² = 2ad

(17.22)²/(2 · 75) = a

a = 1.978 m/s²

Now, we can begin looking at forces.

For an object moving down a ramp experiencing friction and an applied force, we have the forces:

Fκ  = μMgcosθ  = Force due to kinetic friction

Mgsinθ = Force due to gravity

A = Applied Force

We can write out the summation. Let down the incline be positive.

ΣF = A + Mgsinθ - μMgcosθ

Or:

ma = A + Mgsinθ - μMgcosθ

We can plug in the given values:

22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))

A = 46.203 N