Question

A 2150 kg truck is travelling along a straight level road at a constant speed of 55 km/hr. Then the driver removes his foot from the accelerator. After 20 sec, the truck speed is 33 km/hr. What is the magnitude of the average net force acting on the truck during the 20 sec interval?

• A. 650.0 N
• B. 670 N
• C. 656.9 N
• D. 700 N

Answer 1

The correct option is C

656.9 N

Applying, $v=u_0+at$

$33km/hr=55km/hr+a\frac{20}{3600}hr$

Therefore, $a=\frac{(33-55)}{20}3600km/h{r}^2$

$=-3960km/h{r}^2$

Applying, F = ma

$F=2150kg\times (-3960km/h{r}^2)$

=-656.9 N

=656.9 N