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

a) 1.89 m/s b) 172.1 N

Step-by-step 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 = (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{(2*W)/(m)} =\sqrt{(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 = (W)/(d) = (4,475 J)/(26.0m) = 172.1 N`