While driving a 2150 kg car, carl steps on the gas pedal, accelerating at 4.0 m/s2. If the coefficient of friction between his car and the road is
.25, how hard is the engine pulling forward?
a
13867.5 N
7417.5 N
9137.5 N
3332.5 N

d. 3332.5 [N]

Explanation:

To solve this problem we will use newton's second law, which tells us that the sum of forces is equal to the product of mass by acceleration.

Here we have two forces, the force that pushes the car to move forward and the friction force.

The friction force is equal to the product of the normal force by the coefficient of friction.

f = N * μ

f = (m*g) * μ

where:

N = weight of the car = 2150*9.81 = 21091.5 [N]

μ = 0.25

f = (21091.5) * 0.25

f = 5273 [N]

Now as the car is moving forward, the car wheels move clockwise. The friction force between the wheels of the car and the pavement must be counterclockwise, i.e. counterclockwise. Therefore the direction of this force is forward. This way we have:

F + f = m*a

F + 5273 = 2150*4

F = 8600 - 5273

F = 3327 [N]

Therefore the answer is d.

explanation:

newton's third law, for every action there is an equal and opposite reaction.

50 inches should be the resultant vector