Question

A mechanic pushes a 2.60 ✕ 103-kg car from rest to a speed of v, doing 5,430 J of work in the process. During this time, the car moves 29.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.(a) the speed v_____ m/s(b) the horizontal force exerted on the car (Enter the magnitude.)_____ N

Answer 1

Part A. We are given that a car is pushed from rest to a final speed doing 5430 Joules of work. -

To determine the final velocity we will use the work and energy theorem which states that the work done is equal to the change in kinetic energy:

`W=K_f-K_0`

Since the car starts from rest this means that the initial kinetic energy is zero:

`W=K_f`

The kinetic energy is given by:

`K=(1)/(2)mv^2`

Substituting in the formula we get:

`W=(1)/(2)mv_f^2`

Now, we solve for the final velocity. First, we multiply both sides by 2 and divide both sides by the mass:

`(2W)/(m)=v_f^2`

Now, we take the square root to both sides:

`\sqrt{(2W)/(m)}=v_f`

Now, we plug in the values:

`\sqrt{(2(5430J))/(2.6*10^3kg)}=v_f`

Solving the operations:

`2.04(m)/(s)=v_f`

Therefore, the final velocity is 2.04 m/s.

Part B. Now, we use the following formula for work:

`W=Fd`

Where "F" is the force and "d" is the distance.

Now, we set this equation equal to the work done by the force:

`Fd=5430J`

Now, we divide both sides by the distance "d":

`F=(5430J)/(29m)=187.2N`

Therefore, the force exerted is 187.2 Newton.