Acollie drags its bed box across a floor by applying a horizontal force of 6.5 n. the kinetic frictional force acting on the box has magnitude 1.8 n. as the box is dragged through 0.84 m along the way, what are (a) the work done by the collie's applied force and (b) the increase in thermal energy of the bed and floor?

(a) 5.46 J

(b) 1.51 J

Explanation:

Work done=Force* Distance

(a)

Work done= Applied horizontal force * Distance

Substituting distance with 0.84m, Force by 6.5N

Work done=6.5*0.84=5.46 Nm or J

(b)

Increase in thermal energy of the bed and floor

The increase in energy is caused by frictional force whose value is provided as 1.8 N

Therefore, energy=Force*distance

Using distance of 0.84m and force of 1.8N

Energy=1.8*0.84=1.512  Nm or J

lol

explanation:

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:

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