physics harry potter.

Question

Neville Longbottom was dared by his friends to go down Suicide Hill, but at point P he gets a little nervous. From point P to Q he drags a foot and creates some friction, so μk=0.14. Suppose the distance from P to Q is a horizontal 6.7 meters, Neville's mass is 43.5 kg and he has KEP=2892 Joules of kinetic energy at point P. Find his kinetic energy at point Q (KEQ).

Answer 1

Hi Lauren:

The work-(kinetic) energy theorem can also be used here. It states that the net work done on a system equals its change in kinetic energy:

W_net = ΔKE = KE_f - KE_i

So we just have to calculate the work done by each force in the situation described, add them up, and use that to determine the kinetic energy change. 

If only friction is acting along the line of displacement, the work-(kinetic) energy theorem gives:

W_net = W_f = KE_f - KE_i

W_f = F_fdcosα

Since friction always opposes any motion that is happening, it always points directly opposite displacement, and therefore, the angle between force and displacement is always 180^o. So friction ALWAYS does negative work (cos(180^o) = -1).

W_f = -F_fd

F_f = μ_kN

where N = normal force. On a horizontal surface, as described, where only the normal force and gravity act vertically, they must equal each other (since the net vertical acceleration is 0). So, N = mg, and

F_f = μ_kmg = (0.14)(43.5 kg)(9.8 m/s²) = 59.6 N

Then:

W_f = -F_fd = -(59.6 N)(6.7 m) = -399.3 J = KE_f - KE_i

We are told KE_i = 2801 J. So we have:

-399.3 J = KE_f - 2892 J

KE_f = -399.3 + 2892 = 2492.7 J

I hope this helps! Just let me know if you have any questions about this.

Answer 2

The kinetic Energy of a moving body is defined as KE = 1/2 m V² ; therefore the velocity of neville at point p is

 

V_p² = 2 KE_p /m, and V_p =√ 2 KEp/m

substituting V_p = √2(2892 joules)/(43.5kg) = 11.53 m/s

 

as Neville moves from point p to point q his velocity at point q is reduced due to friction.

 

The velocity at point q is governed by the following kinematic relation V_q²-V_p²=2ax, where a

 is the acceleration or deceleration rate

 

From newton Law F = ma, the force F is the force of friction in newtons  

F = μk N = 0.14 x 43.5 kg x 9.81 N/kg = 15.9 N, where 1kg = 9.81 N, and 1 N = kg.m/s

 

F is a negative force since it is a force acting opposite to the direction to motion

 

therefore a = -F/m = -15.9 kg.m/s / 43.5 kg = -0.37 m/s, where a is a deceleration

 

therefore V_q² = -2ax+Vp² = 2 x -0.37 m/s x 6.7 m + (11.53 m/s)² = 128 m²/s² , V_q = 11.31 m/s

 

the kinetic energy at point q is KE_q = 1/2 m Vq² = 1/2(43.5kg)(128m²/s²) = 2784 joules

 

checking KE_q < KE_p , OK