A fireman directs a stream of water into the watertight bed of a truck. The wate

Question

A fireman directs a stream of water into the watertight bed of a truck. The water emerges horizontally from the nozzle at rate 261.0 kg/min with speed 37.50 m/s. Take the water to hit the front of the bed of the truck horizontally and then to drop down and accumulate without splashing out. The mass of the truck (bed empty) is 2000 kg and it is free to move without friction. Initially the truck is at rest and the bed is empty. What is the magnitude of its acceleration at that instant? Two minutes later the truck is moving with a speed of 7.76 m/s. What is its acceleration at that instant? m/s^2 m/s^2

Explanation / Answer

mass flowrate, dm/dt = 261 kg/min

= 261/60 kg/s

= 4.35 kg/s

a) Force exerted by the truck on water = the rate of change on momentum of water

= dP/dt

= v*dm/dt

= 37.5*4.35

= 163 N

according to Newton's third law the same amount force is exerted by water on the truck.

so, acceleration of truck, a = F/M

= 163/2000

= 0.0816 m/s^2

b) after two min, mass of the truck, M = 2000 + 261*2

= 2522 kg

Force exerted by the truck on water = the rate of change on momentum of water

= dP/dt

= v*dm/dt

= (37.5 - 7.76)*4.35

= 129 N

according to Newton's third law the same amount force is exerted by water on the truck.

so, acceleration of truck, a = F/M

= 129/2522

= 0.0513 m/s^2 <<<<<<<<<<<-----------------------Answer

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