two blocks with masses 2 kg and 4 kg are pushed from rest by the same amount of fore for a distance of 100 m on a frictionless floor. the final kinetic energy of the 2 kg block after the 100 m distance is

the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J

Explanation:

mass of block A = 2 kg

mass of block B = 4 kg

distance the blocks were pushed = 100 m

NB: Blocks were pushed the same distance at the same equal time period. And the ground is without friction.

Work done in moving the 2 kg mass along the 100 m distance is,

work = force x distance moved

force exerted by the 2 kg mass = 2 x 9.81 m/s^2(acceleration due to gravity)

force = 19.62 N

therefore,

work done = 19.62 x 100 = 1962 Joules of work.

According to energy conservation principles, the kinetic energy impacted of the 2 kg mass through this distance will be equal to the work done in moving the 2 kg mass through this distance.

Therefore, the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J

gets less

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card