Question

A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the Fig., a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block’s speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?

Answer 1

Given,

Mass (m) = 10.0 kg

Kinetic energy at 0 m = 180 J

(i) Speed at 0 m:

kinetic anergy = `1/2 "mv"^2`

`180 = 1/2 xx 10 xx v^2`

`v^2 = (180 xx 2)/10 = 36`

v = 6 m s^–1

Hence, the speed of the block at 0 m is 6 m s^–1.

(ii) Speed at 4 m:

The area under the force-displacement graph (Fig.) between 0 and 4 meters represents the work that the variable force has done on the block.

Area under the graph = area of left triangle + area of rectangle + area of right triangle

= `1/2 xx 1 xx 50 + 2 xx 50 + 1/2 xx 1 xx 50`

= 25 + 100 + 25 = 150 J

Using the work-energy theorem,

Kinetic energy at 4 m = Kinetic energy at 0 m + Work done

Kinetic energy at 4 m = 180 + 150 = 330 J

Now,

`1/2 "mv"^2 = 330`

`1/2 xx 10 xx v^2 = 330`

v² = 66

`v = sqrt(66) ≈ 8.1 "m s"^-1`

Hence, the speed of the block at 4 m is about 8.1 m s^–1.

Negative acceleration: There is very little friction and the applied force is always in the block's direction of motion. Where the applied force is positive, the acceleration is positive; only when the force is 0 does the acceleration become zero. As a result, there is no negative acceleration throughout any part of the block's motion.