# Jim is driving a 2268-kg pickup truck at 30.0 m/s...?

Jim is driving a 2268-kg pickup truck at 30.0 m/s and releases his foot from the accelerator pedal. The car eventually stops due to an effective friction force that the road, air, and other things exert on the car. The friction force has an average magnitude of 900 N .

I got the first two parts of the question which was

Determine the initial kinetic energy of the truck?

1.02x10^6 J

Determine the stopping distance of the truck.

1130 m

It's this part that im stuck in, i'm not exactly sure what to use to plug into the equation.

Determine the coefficient of kinetic friction between tire and surface:

I know the equation is Fk=uFk*N

But i don't have "N"

I got the first two parts of the question which was

Determine the initial kinetic energy of the truck?

1.02x10^6 J

Determine the stopping distance of the truck.

1130 m

It's this part that im stuck in, i'm not exactly sure what to use to plug into the equation.

Determine the coefficient of kinetic friction between tire and surface:

I know the equation is Fk=uFk*N

But i don't have "N"

## 3 Answers

- They already told you what the average frictional force is. So set the Work done by friction (Work = force x distance) equal to the kinetic energy. The Work is done to absorb all of the kinetic energy. (900 N)(distance) = 1.02 x 10^6 Joules. I get 1133.333 meters01
- Since the friction force causes the truck to decelerate from 30 m/s to 0 m/s, the work that is done by the friction force is equal to the truck’s initial kinetic energy.

KE = ½ * 2268 * 30^2 = 1,020,600 J

This rounds to 1.02 * 10^6 J.

900 * d = 1,020,600

d = 1,020,600 ÷ 900 = 1134 meters.

This rounds to 1130 m01 - N is the normal force of the road against the truck (via its tires). This is, by Newton's 3rd Law, equal to the weight of the truck = mg = (2268)(9.81) = 22,249 N.

Now here's where it get's confusing. The description in this question mentions "effective friction force" which is made up of MORE than the kinetic friction force because air friction and "other things" are included.

So the way it is worded => it's really not possible to compute the coefficient of kinetic friction between the truck's tires and road..because the effective friction force, given, is greater than the kinetic friction force.

=>Ask teacher about to confirm this fact <=

Otherwise if one assumes that the avg friction force = the avg KINETIC friction force, ALONE, then:

µk = 900/22,249 ≈ 0.04 <= way too small for a normal tire/road situation {something ain't right here :>)10

### Trending

- Kinematic eq?
- An rc circuit has a resistancs of 2k ohms and a capacitance of .01592 microfarad. the frequency is 5khz. what is the phase angle?
- Is a gravity field basically a force potential?
- La formula en la que una bateria de automovil y un capacitor almacenan energia?
- What would happen if a nuclear power plant explodes in a small town?
- I am trying to understand forces and momentum, please help. The main thing is, if a e.g. bat knocks a ball, once the bat stops touching the?
- Could railgun rounds lose velocity in space?
- What would happen if we launched all our nuclear weapons at the sun?
- Is it possible for a person to run at the speed of light?
- How can you stop temperature from effecting a resistor in a circuit (so the resistance is more constant and unaffected by temp).?