A wagon with two boxes of gold, having total mass 300 k g , is cut loose from th

Question

A wagon with two boxes of gold, having total mass 300 kg, is cut loose from the horses by an outlaw when the wagon is at rest 50 m up a 6.0 ? slope (see the figure(Figure 1) ). The outlaw plans to have the wagon roll down the slope and across the level ground, and then fall into a canyon where his confederates wait. But in a tree 40 m from the canyon edge wait the Lone Ranger (mass 75.0 kg) and Tonto (mass 60.0 kg). They drop vertically into the wagon as it passes beneath them.

i) When the two heroes drop into the wagon, is the kinetic energy of the system of the heroes plus the wagon conserved?

ii) If not, does it increase or decrease, and by how much? If conserved type 0.

****I know that the answer to part i) is no. How do you solve part ii? THANK YOU

Explanation / Answer

use the conservation of energy principle. When the wagon was at rest, it possessed gravitational potential energy given by

GPE = m*g*h

where m and g are mass and 9.8 m/s^2. The variable h is height -- that wouldn't be 50 m, you need to do some trig to find what the vertical height is.

When the wagon is down on level ground, all the GPE is converted to kinetic energy of the same number of Joules. So the equation is

GPE = KE = (1/2)*m*v^2

Plug in the value of GPE and solve for velocity v. That will be the velocity of the wagon just before the lone ranger and tonto drop in to visit. The velocity will change according to the principle of the conservation of momentum. The total horizontal momentum be the same, before and after they drop in.

Momentum before = momentum after

300 kg*v1 = (300 kg + 75 kg+ 60 kg)*v2

where v1 = v from above and v2 is the velocity after the lone ranger and tonto dropped in. Solve for v2.

Now I see that we ignore friction.

a) Now use v2 in a basic kinematic formula and see how far the wagon goes in 5.0 s.

b) Calculate the new kinetic energy

KEnew = (1/2)*m*(v2)^2

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