Find the change of the mechanical energy of a 1000-kg car traveled down an incline with 1:10 grade a total distance of 100 m during which its speed increased from 10 m/s to 20 m/s?
Is my answer correct, if not what did I do wrong?
ME=KE+PE+Q
KE=[.5*1000*20^2]- [.5*1000*10^2]
KE=200000 - 50000 = 150000
PE=[1000*9.8*10]-[1000*9.8*0]
PE=9800
150000+9800 = 248000J
the kinetic energy increased but the potential energy DECREASED
1000 * 9.8 * 10 = 98,000
To find the change in mechanical energy (ΔME) of the car, you need to take into account the change in kinetic energy (ΔKE) and potential energy (ΔPE) as it travels down the incline. The correct formula is:
ΔME = ΔKE + ΔPE
Let's calculate the values:
1. ΔKE (change in kinetic energy):
You correctly calculated the initial kinetic energy (KEi) as 0.5 * 1000 * 10^2 = 50000 J.
The final kinetic energy (KEf) is given as 0.5 * 1000 * 20^2 = 200000 J.
Therefore, ΔKE = KEf - KEi = 200000 J - 50000 J = 150000 J.
2. ΔPE (change in potential energy):
The change in potential energy is zero because the car starts and ends at the same height. So, ΔPE = 0 J.
Now, summing up ΔKE and ΔPE:
ΔME = ΔKE + ΔPE = 150000 J + 0 J = 150000 J.
Therefore, the change in mechanical energy of the car as it travels down the incline is 150000 J. It seems that your answer of 248000 J is incorrect.