A 2600 kg truck traveling northat 38 km/h turns east andaccelerates to 50 km/h.

Question

A 2600 kg truck traveling northat 38 km/h turns east andaccelerates to 50 km/h. (a) What is the change in the truck's kinetic energy?
J

(b) What is the magnitude of the change in the linear momentum ofthe truck?
kg·m/s

(c) What is the direction of the change in the linear momentum ofthe truck?
°(measured clockwise from east) (a) What is the change in the truck's kinetic energy?
J

(b) What is the magnitude of the change in the linear momentum ofthe truck?
kg·m/s

(c) What is the direction of the change in the linear momentum ofthe truck?
°(measured clockwise from east)

Explanation / Answer

Mass of truck m = 2600 kg a) Change in KE = 0.5 m (vf2 -vi2) = 0.5 * 2600 * (193.20 -110.25 ) =107848 J b) P = m (vf-vi) = 2600 ( 3.4 ) = 8840 kg.m/s c) = tan-1 (vi / vf) =37.23o

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