PLEASE HELP. a 1000 kg car is coasting down the highway at 20 m/s. How much work do friction and air resistance do in bringing the car to rest?
To calculate the work done by friction and air resistance in bringing the car to rest, we need to consider the change in kinetic energy.
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
Given:
Mass of the car (m) = 1000 kg
Initial velocity (v_i) = 20 m/s
Final velocity (v_f) = 0 m/s
We can calculate the initial kinetic energy (KE_i) of the car using the formula:
KE_i = (1/2) * m * v_i^2
Plugging in the values, we get:
KE_i = (1/2) * 1000 kg * (20 m/s)^2
Next, we can calculate the final kinetic energy (KE_f) since the car comes to rest:
KE_f = (1/2) * m * v_f^2
Plugging in the values, we get:
KE_f = (1/2) * 1000 kg * (0 m/s)^2 = 0
The change in kinetic energy (ΔKE) is then given by:
ΔKE = KE_f - KE_i = 0 - [(1/2) * 1000 kg * (20 m/s)^2]
Simplifying further, we have:
ΔKE = - (1/2) * 1000 kg * (400 m^2/s^2)
Now, since the work done by friction and air resistance is equal to the change in kinetic energy, we can calculate it.
Work = ΔKE = - (1/2) * 1000 kg * (400 m^2/s^2)
Performing the calculation, we get the answer:
Work = - (1/2) * 1000 kg * (400 m^2/s^2)
= - 200,000 J
Therefore, the work done by friction and air resistance in bringing the car to rest is -200,000 Joules (J). The negative sign indicates that work is done against the motion, resulting in a loss of energy.