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:

$K=\frac{1}{2}mv^2$

Substituting in the formula we get:

$W=\frac{1}{2}mv_f^2$

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

$\frac{2W}{m}=v_f^2$

Now, we take the square root to both sides:

$\sqrt{\frac{2W}{m}}=v_f$

Now, we plug in the values:

$\sqrt{\frac{2(5430J)}{2.6\times10^3kg}}=v_f$

Solving the operations:

$2.04\frac{m}{s}=v_f$

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":

$F=\frac{5430J}{29m}=187.2N$

Therefore, the force exerted is 187.2 Newton.