Conservation of Energy EX 3

(a) The initial mechanical energy is the sum of the kinetic energy and the gravitational potential energy:

$E=K+U_{G}$

$E={\frac {1}{2}}mv^{2}+mgh$

$E={\frac {1}{2}}(3.2)(5)^{2}+(3.2)(9.8)(4)$

$E=40+125.44=165.44J$

The gravitational potential energy at a height of 5 m is:

$U_{G}=(3.2)(9.8)(5)=156.8J$

This means that the trolley has enough energy to climb up to 5 m so the trolley does reach the spring.

(b) Since there are no nonconservative forces, the mechanical energy is always 156.8 J. When the spring is compressed, there is no kinetic energy. The mechanical energy is the sum of the spring potential energy and the gravitational potential energy:

$E=U_{K}+U_{G}$

$156.8={\frac {1}{2}}kx^{2}+mgh$

$156.8={\frac {1}{2}}(120)x^{2}+(3.2)(9.8)(2)$

$156.8=60x^{2}+62.72$

$94.08=60x^{2}$

$x^{2}=1.56$

$x=1.3m$

Alternative solution:

(a) To reach the spring, the trolley must still have some kinetic energy when it is at the top of the hill (5 m). The trolley is the object of our study. It is initially 4 m above ground. At the end, the trolley is at 5 m and we want to know whether its velocity is positive.

1. The change in kinetic energy:
   $\Delta K=K_{f}-K_{i}=K_{f}-{\frac {1}{2}}(3.2)(5)^{2}=K_{f}-40$
   
   
2. The change in potential energy:
   The spring does not change during this event and so we will only compute the gravitational potential energy.
   $\Delta U_{G}=mgh_{f}-mgh_{i}=(3.2)(10)(5)-(3.2)(10)(4)=32J$
   
   
3. Work done by non-conservative forces is zero.
   
   
4. Substitute into the equation for conservation of energy:
   $\Delta K+\Delta U_{G}=K_{f}-40+32=0$
   $K_{f}=8J$
   
   
5. The solution indicates that the kinetic energy is positive at the end and so the trolley moves over the hump.

(b) The maximum compression indicates that at the end the trolley is brought to rest (hence, no more compression). At the beginning, the trolley is at the height of 4 m, moves with a speed of 5 m/s and the spring's compression is zero. At the end, the trolley is at the height of 2 m, the spring is compressed and the trolley does not move towards the right anymore.

1. The change in kinetic energy:
   $\Delta K=K_{f}-K-i=0-40=-40J$
   
   
2. The change in potential energy:
   $\Delta U_{G}=mgh_{f}-mgh_{i}=(3.2)(10)(2)-(3.2)(10)(4)=-64J$
   $\Delta U_{sp}={\frac {1}{2}}k{x_{f}}^{2}-{\frac {1}{2}}k{x_{i}}^{2}={\frac {1}{2}}(120){x_{f}}^{2}-0={\frac {1}{2}}(120){x_{f}}^{2}$
   $\Delta U=\Delta U_{G}+\Delta U_{sp}=-64+{\frac {1}{2}}(120){x_{f}}^{2}$
   
   
3. No work is done by non-conservative forces.
   
   
4. Substitute into the equation for conservation of energy and solve:
   $\Delta K+\Delta U=-40+(-64+{\frac {1}{2}}(120){x_{f}}^{2})=0$
   ${\frac {1}{2}}(120){x_{f}}^{2}=108$
   $x_{f}={\sqrt {108 \over 60}}=1.3m$