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I need to find Wfg (Work by friction) ,Wnet (net Work) the acceleration of the car and how long it will take Tyler to get to the gas station

I need to find Wfg (Work by friction) ,Wnet (net Work) the acceleration of the car-example-1
User Arsal
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Step-by-step explanation:

First, let's calculate the work done by the force applied by Taylor.

Wt = Fd = (1.3 x 10⁴ N)(15 m) = 195000 J

Because the work is the force times the distance.

Now, to calculate the force of friction, we get:

Ff = (Fn)μ = mgμ

Ff = (1200kg)(9.8 m/s²)(0.67)

Ff = 7879.2 N

Where Fn is the normal force and it is equal to the weight of the car. m is the mass and g is the acceleration due to gravity.

Then, the work done by the friction is

Wf = Ff d = (7879.2 N)(15 m) = 118188 J

Therefore, the net work is equal to

Wnet = Wt - Wf

Wnet = 195000J - 118188 J

Wnet = 76812 J

To find the acceleration, we need to find the net Force

Fnet = F - Ff

Fnet = 13000N - 7879.2N

Fnet = 5120.8 N

So, by the second law of Newton, the acceleration is equal to

a = Fnet/m

a = 5120.8 N / 1200 kg

a = 4.27 m/s²

Finally, by the kinematics equations, we get:

d = vt + (1/2)at²

Since it starts from rest, we get:

d = (1/2)at²

Solving for t, we get:

2d = at²

2d/a = t²

t = √(2d/a)

So, replacing d

User Philipp Zedler
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