Tom has his friend Steve give him an initial push uphill. After the initial push, Steve just let's Tom "coast" up the hill

Question

Steve pushes the sled with an initial speed of 3.7 m/s up a 33° incline. The combined mass of Tom and the sled is 19.2 kg, and the coefficient of kinetic friction between the sled and the incline is 0.20. How far up the incline does the sled move?

Answer 1

Hi Charlotte A.!

To answer this question, we should start with a free body diagram. It is most efficient if we tilt the axes of our plane so that it lines up with the normal force on the y-axis and the force due to friction along the negative x-axis. In this case, our force due to gravity will point at angle of 33° left of the normal into the third quadrant.

Once we have all of our forces drawn and labelled, we can make good use of the conservation of energy:

E_o + W_nc = E_f

Now, we must considered which energies we have at the initial and final positions, as well as whether we have any non-conservative forces doing work.

Initially we are on ground level, where h = 0m. We are also moving. So, we have no initial gravitational potential energy, but we do have initial kinetic energy.

In the final position, we assume the sled comes to a rest, so we are no longer moving. However, we have moved up against the force due to gravity. So, we have no kinetic energy, but we have gravitational potential energy.

We are given a coefficient of kinetic friction. So, we must consider the dissipative work done by friction.

We are ready to put all of this into our equation:

KE_o + W_f = Ug_f

(1/2)mv_o² + (µ_kn)Lcos(180°) = mgh

The force due to friction points in the opposite direction to movement. So, the dissipative work done is likewise in the opposite direction. In this case, we take the angle between the force due to friction and the direction as 180°:

cos180° = -1

Because we are at an incline, the normal force is not simply equal to the weight of sled. We must consider the angle as well.

n = mgcosθ

We are not given the height of this hill. So, we can substitute in other terms that are more valuable to us:

sinθ = h/L => h = Lsinθ

Putting this together:

(1/2)mv_o² + (µ_kn)Lcos(180°) = mgh

=> cos180° = -1

=> n = mgcosθ

=> sinθ = h/L => h = Lsinθ

(1/2)mv_o² - (µ_kmgcosθ)L = mgLsinθ

Note that the mass cancels:

(1/2)v_o² - (µ_kgcosθ)L = gLsinθ

Now, we simply rearrange to solve for L:

(1/2)v_o² = gLsinθ + (µ_kgcosθ)L = L[gsinθ + (µ_kgcosθ)]

L = (1/2)v_o²/[gsinθ + (µ_kgcosθ)] = v_o²/2g[sinθ + (µ_kcosθ)]

<=> (3.7 m*s^-1)²/[2*9.8 m*s^-2(sin33° + 0.2*cos33°)]

L = 0.98m

Hope this helps!

Cheers