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Conservation of Energy

Practice Conservation of Energy
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Conservation of Energy

Energy is conserved during a roller coaster ride

There are many energy conversions between potential and kinetic energy as the cars travel around this double looping roller coaster. Throughout the ride, however, there will always be the same total amount of energy.

Conservation of Energy




PE = mgh = (20.0 \ N)(2.00 \ m) = 40.0 \ J

If the box slides off the shelf, the only force acting on the box is the force of gravity and so the box falls. We can calculate the speed of the box when it strikes the ground by several methods.  We can calculate the speed directly using the formula v{_f}^2 = 2ad .  We can also find the final velocity by setting the kinetic energy at the bottom of the fall equal to the potential energy at the top,  KE = PE , thus  \frac{1}{2} \ mv^2 = mgh . When reduced, we see that  v{_f}^2 = 2gh . Note that these formulas are essentially the same; when gravity is the acceleration and the height is the distance, they are the same equation.

v=\sqrt{(2)(9.80 \ m/s^2)(2.00 \ m)}=6.26 \ m/s



E_{\text{TOTAL}} = KE + PE = \frac{1}{2} \ mv^2 + mgh

=\left(\frac{1}{2}\right)(5.00 \ kg)(30.0 \ m/s)^2 + (5.00 \ kg)(9.80 \ m/s^2)(50.0 \ m)

= 2250 \ J + 2450 \ J = 4700 \ J

\frac{1}{2} \ mv{_f}^2 = 4700 \ J \

\ v_f=\sqrt{\frac{(2)(4700 \ J)}{5.00 \ kg}}=43.4 \ m/s




(c) How much  KE was lost in the collision?


(a)  KE_{\text{BULLET}} = \frac{1}{2} \ mv^2 = \left(\frac{1}{2}\right)(0.00200 \ kg)(705 \ m/s)^2 = 497 \ J

(b) m_Bv_B + m_Wv_W = (m_{B+W})(v_{B+W})

(0.00200 \ kg)(705 \ m/s) + (0.250 \ kg)(0 \ m/s) = (0.252 \ kg)(V)

(1.41 \ kg \ m/s) =(0.252 \ kg)(V)

V = 5.60 \ m/s

(c) KE_{\text{COMBINATION}} = \frac{1}{2} \ mv^2 = \left(\frac{1}{2}\right)(0.252 \ kg)(5.60 \ m/s)^2 = 3.95 \ J

KE_{\text{LOST}} = KE_{\text{BEFORE}} - KE_{\text{AFTER}} = 497 \ J - 4 \ J = 493 \ J





The following video demonstrates Newton Ball tricks. Use this resource to answer the questions that follow.


  2. What h appens when three balls are pulled up to one side and released?

Practice problems with answers for the law of conservation of energy:




    2. what is its velocity?
  4. A skier starts from rest at the top of a 45.0 m hill, coasts down a  30^\circ slope into a valley, and continues up to the top of a 40.0 m hill.  Both hill heights are measured from the valley floor.  Assume the skier puts no effort into the motion (always coasting) and there is no friction.

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