An iron block of mass m = 500 kg is kept at the back of a truck moving at a speed `v_(0)=90km h^(-1)`. The driver applies the brakes and shows down to a speed of `v=54km h^(-1)` in 10s. What constant force acts on the block during this time if the block does not slide on the truck-bed?

Explanation

The acceleration, `a=(v-v_(0))/(t)`
`=(54 km h^(-1)-90kmh^(-1))/(10s)`
`=(15ms^(-1)-25ms^(-1))/(10s)`
`=-1m//s^(2)`
or, Force F = ma
`=500kg (-1 ms^(-2))`
`=-500N`
-ve sign indicates that the force acts opposite to the velocity of the block. The magnitude of force is 500 N.