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:
[tex]\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J[/tex]
- where m= mass of the truck = 2.5*10³ kg.
- So, we can find the speed v, as follows:
[tex]v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s[/tex]
b)
- The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:
[tex]W = F*d[/tex]
- We can solve for F, as follows:
[tex]F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N[/tex]
