Marissa drags a 15 kg duffel bag 18 m across the gym floor. If the coefficient of kinetic friction between the floor and bag is 0.15, how much thermal energy does Marissa create?
It says I'm look for deltaE_Th. I don't know what that means or how to figure out the problem.
Never mind I got it!
To find the thermal energy (ΔE_Th) that Marissa creates while dragging the duffel bag, you can use the equation:
ΔE_Th = μ * m * g * d
Where:
ΔE_Th is the thermal energy generated (in joules)
μ is the coefficient of kinetic friction (0.15 in this case)
m is the mass of the duffel bag (15 kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
d is the distance the duffel bag is dragged (18 m)
Now let's calculate the thermal energy:
ΔE_Th = 0.15 * 15 kg * 9.8 m/s^2 * 18 m
First, multiply 0.15 by 15 kg:
ΔE_Th = 2.25 * 9.8 m/s^2 * 18 m
Next, multiply 2.25 by 9.8:
ΔE_Th ≈ 22.05 m/s^2 * 18 m
Now, multiply 22.05 by 18:
ΔE_Th ≈ 396.9 J
Therefore, Marissa creates approximately 396.9 joules of thermal energy while dragging the duffel bag across the gym floor.