A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the given figure 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:

(A) at 0 m, and
(B) at 4 m.

Does the block have negative acceleration in any portion of its motion?

Given:
Mass of block, m = 10 kg
Initial kinetic energy at 0 m:
K.E. = 180 J
(A) Speed at 0 m:

$$\text{Using, } K.E. = \frac{1}{2}mv^2$$

$$180 = \frac{1}{2} \times 10 \times v^2$$

$$180 = 5v^2$$

$$v^2 = 36$$

$$v = 6 \text{ ms}^{-1}$$

Therefore, the speed of the block at 0 m is 6 ms$^{-1}$

(B) Speed at 4 m:

Work done by the force = Area under the force-displacement graph.

From 0 m to 1 m:

$$\text{Area of triangle} = \frac{1}{2} \times 1 \times 50$$

$$= 25 \text{ J}$$

From 1 m to 3 m:

$$\text{Area of rectangle} = 2 \times 50$$

$$= 100 \text{ J}$$

From 3 m to 4 m:

Area of triangle

$$= \frac{1}{2} \times 1 \times 50$$

$$= 25 \text{ J}$$

Total work done:

$$W = 25 + 100 + 25$$

$$W = 150 \text{ J}$$

Using work-energy theorem,

$$\text{Final K.E.} = \text{Initial K.E.} + W$$

$$= 180 + 150$$

$$= 330 \text{ J}$$

Now,

$$330 = \frac{1}{2} \times 10 \times v^2$$

$$330 = 5v^2$$

$$v^2 = 66$$

$$v = \sqrt{66}$$

$$v\thickapprox8.1\text{ ms}^{-1}$$

Therefore, the speed of the block at 4 m is: 8.1 ms$^{-1}$.
The block does not have negative acceleration during any part of its motion because the applied force is always positive and acts in the direction of motion.