A block of mass $1\,\text{kg}$ lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is $0.6$. If the acceleration of the truck is $5\,\text{m/s}^{2}$, the frictional force acting on the block is ..... newtons.

Step 1: Identify the given parameters.

The mass of the block, the coefficient of static friction, and the acceleration of the truck are given.

$$m = 1\,\text{kg}$$ $$\mu_s = 0.6$$ $$a = 5\,\text{m/s}^{2}$$

Step 2: Determine the normal force acting on the block.

In the vertical direction, the normal force balances the gravitational force.

$$N = mg$$

Using $g = 9.8\,\text{m/s}^{2}$:

$$N = 1 \times 9.8 = 9.8\,\text{N}$$

Step 3: Calculate the maximum static frictional force.

The maximum static friction that the surface can provide is determined by the normal force.

$$f_{\max} = \mu_s N$$ $$f_{\max} = 0.6 \times 9.8 = 5.88\,\text{N}$$

Step 4: Calculate the force required to accelerate the block with the truck.

For the block to move together with the truck without slipping, the static frictional force must provide the necessary acceleration.

$$f_{\text{req}} = ma$$ $$f_{\text{req}} = 1 \times 5 = 5\,\text{N}$$

Step 5: Compare the required force with the maximum static friction.

Since the required force is less than the maximum available static friction, the block does not slip relative to the truck.

$$5\,\text{N} < 5.88\,\text{N}$$

Therefore, the actual static frictional force acting on the block is equal to the required force.

$$f = 5\,\text{N}$$

Answer: 5