Question

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m. Neglecting friction between truck and road, determine the following:
a. The speed v.
b. The horizontal force exerted on the truck.

Answer 1

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

• Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
• This work, is just 4,475 J.
• So we can write the following equation:

        $\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

• where m= mass of the truck = 2.5*10³ kg.
• So, we can find the speed v, as follows:

        $v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

b)

• The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        $W = F*d$

• We can solve for F, as follows:

        $F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$